Statistics · S.IC.B.4

One sample interval for a mean and sample size

A random sample of animals has been weighed. What range of values for the mean of the whole herd does the sample support, and how many animals must be weighed to pin that mean down to within a chosen margin?

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Conditions for one meanStatistics Critical valueStatistics The t distribution and degrees of freedomStatistics

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Interval for a difference of meansStatistics Interval for a slopeStatistics Paired data: interval and test for a mean differenceStatistics

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One sample interval for a mean and sample size

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When you reach for it

Illari runs a llama ranch in the highlands of Peru, and every year each animal is shorn and its fleece weighed (made-up data throughout this lesson). One herd has 600 llamas. The wool buyer wants an estimate of \mu, the mean fleece weight for the whole herd this year, before shearing is finished, so Illari shears a simple random sample of 15 llamas first. Their mean fleece is 4.2 pounds. The buyer does not care about these 15 animals; the price rests on all 600, and another 15 would have given another mean.

The one-sample interval for a mean turns one sample mean into a range of plausible values for the population mean, with a stated confidence level. It is the move for any measured quantity in one random sample: weights, times, lengths, scores. Its second use comes before any data exist: run backwards, the same formula says how many individuals to measure so that the interval comes out no wider than wanted.

The parts are already in place. Conditions for one mean says when the sample mean behaves well enough for this: a random sample, no more than a tenth of the population, and a population close to normal or a sample large enough. The t distribution and degrees of freedom supplies the curve to use when the population’s spread has to be estimated from the sample, with n - 1 degrees of freedom. Critical value turns a level into a multiplier. The same plan for a share instead of a mean is One sample interval for a proportion and sample size.

The procedure

Day 1: the interval

Every interval of this family has the form “estimate, plus or minus a critical value times a standard error.” For a mean the estimate is \bar{x} and its Standard error is \displaystyle s_x / \sqrt{n}, which gives

\bar{x} \pm t^* \frac{s_x}{\sqrt{n}}, \qquad \text{df} = n - 1

The amount added and subtracted is the Point estimates and the margin of error.

Why t^* and not z^*. If the population standard deviation \sigma were known, \displaystyle \dfrac{\bar{x} - \mu}{\sigma / \sqrt{n}} would follow the standard normal curve and z^* would be right. It is not known, so s_x stands in for it, and s_x changes from sample to sample as well. That extra wobble makes \displaystyle \dfrac{\bar{x} - \mu}{s_x / \sqrt{n}} follow a t distribution with n - 1 degrees of freedom, whose tails are heavier. The critical value t^* is the number with the chosen central area of that curve between -t^* and t^*. It is larger than z^*, and the extra width is exactly what keeps the capture rate at the stated level. With n = 15, an interval built with z^* = 1.96 would capture \mu in only about 93\% of samples, not 95\%.

Name the parameter first. \mu is the mean of a stated variable over a stated population: “the mean fleece weight, in pounds, of all 600 llamas in the herd this year.” It is never the mean of the sample.

Check the conditions, briefly but every time. A random sample makes \bar{x} center on \mu. A sample no more than a tenth of the population keeps the draws close to independent, which the standard error assumes. Then the shape: with n \ge 30 the sampling distribution of \bar{x} is close to normal unless the data are very strongly skewed; with a smaller sample, a graph of the data must show no strong skew and no outliers. The reasons are in Conditions for one mean. When all three hold, the procedure is called a one-sample t interval.

Find t^*. In a t table, go to the row for the degrees of freedom and the column for the level. When the exact row is missing, use the next smaller degrees of freedom: its t^* is a little larger, so the interval is a little wider, which errs on the safe side. A calculator’s invT command, or a spreadsheet’s `T.INV`, takes the area to the left of t^* and the degrees of freedom; for 95\% that area is 0.975.

Compute and conclude in context. The sentence names the level, both ends with units, the population and the variable: “We are 95\% confident that the mean ... of all ... is between ... and ...”. What the 95\% refers to, the method across all possible samples, is the subject of Confidence level and interpreting an interval.

Technology. A calculator’s TInterval takes either the data list or the three numbers \bar{x}, s_x, n, and returns the two ends. In a spreadsheet, `CONFIDENCE.T(0.05, s, n)` returns the 95\% margin directly. Neither checks a condition; the checks stay with the person running it.

Changing the level changes only t^*. The center \bar{x} and the standard error stay put, so a 90\% interval sits inside the 95\% one, which sits inside the 99\% one.

Day 2: choosing a sample size, and judging a claim

Before the data exist, the margin a sample of size n will give is about \displaystyle z^* \sigma / \sqrt{n}. Asking for a margin no larger than m and solving for n (both sides are positive, so squaring keeps the inequality):

n \ge \left(\frac{z^* \sigma}{m}\right)^2

Two substitutions are made here, and both should be said out loud. First, \sigma is unknown, so a planning value stands in for it: the s_x of an earlier sample, a pilot study, or past records, often rounded up to be safe. Second, the real interval will use t^*, not z^*, but t^* depends on the degrees of freedom, which depend on the n being solved for. Using z^* breaks that circle. Because t^* is always a little larger than z^*, the n from the formula is a slight underestimate: if s_x comes out equal to the planning value, the margin lands just above m. Adding a few individuals, or checking the margin with the t^* for the chosen n and stepping up, closes the gap.

The formula usually returns a decimal. The sample size must be a whole number at least that large, so it is rounded up, always. The population size does not appear: as long as the planned sample stays under a tenth of the population, a herd ten times as large needs the same sample for the same margin.

Judging a claimed mean. An interval is the set of values of \mu that the sample supports at its level. A claimed value outside the interval is evidence against the claim at that level; a value inside is plausible, which is not the same as shown to be true. The One sample test for a mean asks the same question with a probability attached, and a two-sided test at the 0.05 level rejects exactly the claimed values that lie outside the 95\% interval.

Worked examples

Example 1

Day 1: the fleece interval

Illari’s random sample of 15 llamas from the herd of 600 gave these fleece weights, in pounds:

\begin{aligned} &2.6,\ 3.0,\ 3.2,\ 3.5,\ 3.7,\ 3.9,\ 4.1,\ 4.2, \\ &4.4,\ 4.6,\ 4.8,\ 5.0,\ 5.2,\ 5.4,\ 5.4 \end{aligned}

Construct and interpret a 95\% confidence interval for the herd’s mean fleece weight.

Figure 1Illari’s random sample of 15 fleece weights from a herd of 600 llamas (made-up data). No strong skew and no outliers.

Solution

Parameter. Let \mu be the mean fleece weight, in pounds, of all 600 llamas in the herd this year.

Conditions. The 15 llamas were chosen at random from the herd, so the random condition holds. A tenth of 600 is 60, and 15 \le 60, so the draws are close to independent. The sample is smaller than 30, so the data must be looked at. Figure 1 shows the values spread fairly evenly with no strong skew. For outliers, the quartiles are Q_1 = 3.5 and Q_3 = 5.0 (the medians of the lowest seven and the highest seven values), so the interquartile range is 1.5 and the fences are 3.5 - 1.5 \times 1.5 = 1.25 and 5.0 + 1.5 \times 1.5 = 7.25. Every value lies between them. All three conditions hold.

Compute. The fifteen values add to 63.0, so \bar{x} = 4.2 pounds, and their standard deviation is s_x \approx 0.8832 pound. With \text{df} = 15 - 1 = 14, the table gives t^* = 2.145 at 95\%.

\begin{aligned} \text{SE}_{\bar{x}} &= \frac{0.8832}{\sqrt{15}} \approx 0.2280 \\ t^*\,\text{SE}_{\bar{x}} &\approx 2.145 \times 0.2280 \approx 0.489 \end{aligned}

The interval is 4.2 \pm 0.489, from 3.711 to 4.689 pounds.

Conclude. We are 95\% confident that the mean fleece weight of all 600 llamas in the herd this year is between 3.711 and 4.689 pounds.

TInterval run on the list prints (3.7109,\ 4.6891) with \bar{x} = 4.2, s_x = 0.88318 and n = 15, the same interval to the digits shown.

Example 2

Day 1: other levels, read from the output

Amaya, who keeps the ranch’s records, reports 90\% and 99\% intervals as well. Running TInterval on the same list gives these two outputs, each also printing \bar{x} = 4.2, s_x = 0.88318 and n = 15.

Level	Interval
0.90	(3.7984,\ 4.6016)
0.99	(3.5212,\ 4.8788)

Find each margin of error, the t^* the calculator used, and say how much wider the 99\% interval is than the 95\% one.

Solution

The interval is centered at \bar{x}, so each margin is half the width:

\begin{aligned} \tfrac{1}{2}(4.6016 - 3.7984) &= 0.4016 \\ \tfrac{1}{2}(4.8788 - 3.5212) &= 0.6788 \end{aligned}

Dividing each by the standard error 0.2280 recovers the multiplier: 0.4016 / 0.2280 \approx 1.761 and 0.6788 / 0.2280 \approx 2.977, which are the table’s t^* values for 14 degrees of freedom at 90\% and 99\%. Nothing else changed: same center, same standard error.

The 99\% margin is \displaystyle \frac{0.6788}{0.489} \approx 1.388 times the 95\% margin, the same as \displaystyle \frac{2.977}{2.145}, because the margin is t^* times a fixed standard error. Figure 2 shows the three intervals nested. In words: we are 90\% confident that the herd’s mean fleece weight is between 3.798 and 4.602 pounds, and 99\% confident that it is between 3.521 and 4.879 pounds.

Example 3

Day 1: a larger sample, summary statistics only

Tupac manages a second herd of about 900 llamas. A random sample of 40 of them has mean fleece weight \bar{x} = 3.94 pounds and s_x = 1.05 pounds; a dot plot of the 40 values is mildly skewed right with no outliers. Construct a 95\% interval for that herd’s mean, first with invT and then with a table that has rows for 30 and 40 degrees of freedom but not 39.

Solution

Parameter. \mu is the mean fleece weight of all llamas in Tupac’s herd this year.

Conditions. Random sample; 40 \le 90, a tenth of the herd; and n = 40 is at least 30 with only mild skew, so the sampling distribution of \bar{x} is close to normal.

Compute. \displaystyle \text{SE}_{\bar{x}} = 1.05 / \sqrt{40} \approx 0.1660, and \text{df} = 39. The command invT(0.975,\ 39) returns t^* \approx 2.023:

\begin{aligned} 2.023 \times 0.1660 &\approx 0.336 \\ 3.94 \pm 0.336 &\to 3.604 \text{ to } 4.276 \end{aligned}

With only the table, the row for 39 is missing, and the next smaller row is 30, with t^* = 2.042. That gives a margin of about 0.339 and the interval 3.601 to 4.279, a hair wider and so a hair safer. (The row for 40 would give a t^* slightly too small.)

Conclude. We are 95\% confident that the mean fleece weight in Tupac’s herd is between 3.604 and 4.276 pounds.

For comparison, z^* = 1.96 would have given a margin of about 0.325. At 39 degrees of freedom the t curve is already close to normal, so the difference is small, but it does not vanish, and t^* remains the correct multiplier whenever s_x stands in for \sigma.

Example 4

Day 2: a margin of a quarter pound

Next year the buyer wants a 95\% interval for the herd’s mean with a margin of error no larger than 0.25 pound. This year’s sample had s_x \approx 0.88, so Illari plans with \sigma = 0.9. How many llamas should she sample?

Solution

With z^* = 1.96, \sigma = 0.9 and m = 0.25:

\begin{aligned} n &\ge \left(\frac{1.96 \times 0.9}{0.25}\right)^2 \\ &= 7.056^2 \approx 49.79 \end{aligned}

The sample must be a whole number at least 49.79, so the formula says 50 llamas.

Now the honest check. The interval she will compute uses t^* with 49 degrees of freedom, which is 2.010, not 1.96. If next year’s s_x comes out exactly 0.9, the margin is \displaystyle 2.010 \times 0.9 / \sqrt{50} \approx 0.256, just above the target. Stepping up, n = 52 gives about 0.2506, still over, and n = 53 gives about 0.248, under. So the z^* formula fell short by three llamas. The second source of error is larger and cannot be removed: s_x itself will vary, and a sample with s_x above 0.9 gives a wider margin whatever n is. Rounding the planning value up, as Illari did, is the usual protection. Either size passes the tenth-of-the-herd check, since 53 \le 60.

Example 5

Day 2: a smaller margin, a larger guess

Illari asks two questions. (a) How many llamas give a 95\% margin of 0.15 pound, with \sigma = 0.9? (b) If the planning value should really have been \sigma = 1.1, how many give a margin of 0.25?

Solution

(a) The same formula with m = 0.15:

n \ge \left(\frac{1.96 \times 0.9}{0.15}\right)^2 = 11.76^2 \approx 138.30

so 139 llamas. The margin sits in the denominator, squared, so shrinking it from 0.25 to 0.15 multiplies the sample by \displaystyle \left(\frac{0.25}{0.15}\right)^2 = \frac{25}{9} \approx 2.78, and indeed \displaystyle 49.79 \times \frac{25}{9} \approx 138.3. But 139 is more than 60, a tenth of the herd; it is about 23\% of all 600 llamas. Sampling that large a share without replacement makes \bar{x} vary less than \displaystyle s_x / \sqrt{n} says, so the usual interval would be wider than it needs to be. A corrected formula for that case exists and lies beyond this lesson.

(b) With \sigma = 1.1:

n \ge \left(\frac{1.96 \times 1.1}{0.25}\right)^2 = 8.624^2 \approx 74.37

so 75 llamas, against 50 with \sigma = 0.9. The sample size grows with the square of \sigma: \displaystyle \left(\frac{1.1}{0.9}\right)^2 \approx 1.49, about half as many llamas again. This size, too, is more than 60, a tenth of the herd. A planning value that is too small is the costly mistake, because the margin it promised will not be met.

Example 6

Day 2: judging two claims

The buyer says the herd’s fleeces average 5.0 pounds. Last year’s records gave a herd mean of 4.0 pounds. Using the three intervals from this year’s sample, which values are plausible for \mu? Does the sample show the mean has changed since last year?

Figure 2The 90\%, 95\% and 99\% intervals from the same 15 fleeces share one center and grow with the level. Last year’s 4.0 pounds lies inside all three; the buyer’s 5.0 lies outside all three.

Solution

The buyer’s 5.0 lies above the upper end of the 95\% interval, 4.689, by 5.0 - 4.689 = 0.311 pound, and above the 99\% upper end, 4.879, as well. At either level the sample is evidence against the claim: the herd’s mean fleece is plausibly less than 5.0 pounds, and the whole interval lies below it.

Last year’s 4.0 lies inside all three intervals, so it is a plausible value. That does not show the mean is unchanged: every value from 3.711 to 4.689 is plausible at 95\%, including 4.5, which would be a rise. The interval gives no evidence of a change and no evidence against one. If the same llamas were weighed both years, the right tool would work with each animal’s change, as in Paired data: interval and test for a mean difference; comparing two separate samples needs Interval for a difference of means.

Boundaries

The interval needs a random sample from the population named in the parameter, or a randomized experiment. Without one, the formula still produces numbers, but neither the center nor the level means anything about \mu.
The standard error treats the draws as independent, which sampling without replacement allows only when n is at most a tenth of the population.
The t multiplier assumes the sampling distribution of \bar{x} is close to normal. A small sample from a strongly skewed population, or one with outliers, breaks that: an outlier moves \bar{x} and inflates s_x at once, and the true capture rate can differ from the stated level. Larger samples tolerate more skew; no sample size rescues a sample that is not random.
The interval is about the mean, not about individuals. “95\% confident that \mu is between 3.711 and 4.689” says nothing about the share of single fleeces in that range, which is much smaller.
The sample-size formula uses z^* and a planning value for \sigma. Its answer is a slight underestimate even when the planning value is right, and only as good as that value when it is not.
This is the interval for one population mean. Two groups need a difference of means, paired measurements need the paired procedure, and a share needs a different standard error and z^*.

Where this goes next

The One sample test for a mean takes a claimed value such as the buyer’s 5.0 and asks how surprising this sample would be if it were true. Paired data: interval and test for a mean difference reduces before-and-after measurements on the same individuals to one column of differences and then runs exactly this interval on it. Interval for a difference of means keeps the same shape, estimate plus or minus t^* times a standard error, to compare two herds. Regression uses it once more in the Interval for a slope.

Exercises

All data are made up. Find t^* from a t table or with invT; the answers use t^* rounded to three decimal places.

Day 1: carrying out the interval

A random sample of 10 llamas from Tupac’s herd of about 900 gave fleece weights, in pounds, of 3.1, 3.6, 3.8, 4.0, 4.3, 4.5, 4.7, 4.9, 5.2, 5.6. Name the parameter, check the conditions, and construct and interpret a 95\% interval.
A random sample of 20 of the ranch’s roughly 400 yearling llamas, shorn for the first time, has mean fleece \bar{x} = 3.85 pounds and s_x = 0.72 pound; the data are roughly symmetric with no outliers. Construct and interpret a 90\% interval.
A random sample of 25 llamas from Tupac’s herd has mean shearing time \bar{x} = 4.6 minutes and s_x = 1.1 minutes, with no strong skew. Construct a 99\% interval for the herd’s mean shearing time.
A TInterval output reads: interval (4.1303,\ 4.7697), \bar{x} = 4.45, s_x = 0.6, n = 16. Find the margin of error, the standard error, the t^* used, and the confidence level.
For each, say whether the one-sample t interval may be used, and if not, which condition fails. (a) A random sample of 8 llamas, seven of whose fleeces weigh between 3.5 and 4.8 pounds and one 9.0 pounds. (b) A random sample of 45 llamas from a herd of 900, fleece weights mildly skewed right, no outliers. (c) The 12 llamas Amaya could catch most easily in the corral. (d) A random sample of 90 llamas from a herd of 600.
Amaya has a random sample of 6 fleeces with \bar{x} = 4.0 pounds and s_x = 0.8 pound, from a population close to normal, and builds a 95\% interval with z^* = 1.96. Compute her interval and the correct one, and say which is right and why.
A spreadsheet cell holds `=T.INV(0.95, 19)` and another `=T.INV(0.995, 24)`. Give each value, and say what confidence level and sample size each serves.

Day 2: sample size and claims

Find the sample size for a 90\% interval with margin at most 0.2 pound, using \sigma = 0.9.
For the whole ranch of 3{,}000 llamas, find the sample size for a 95\% margin of at most 0.1 pound with \sigma = 0.8, and check the tenth-of-the-population condition.
Tupac has no earlier sample. He knows fleeces run from about 2 to 6 pounds and takes a quarter of that range as his planning value for \sigma. Find the sample size for a 95\% margin of at most 0.3 pound.
The budget covers shearing 30 llamas early. With \sigma = 0.9, what 95\% margin does the z^* formula predict, and what margin results if s_x comes out 0.9 and the interval uses t^*?
For the sample size in Exercise 8, find the actual 90\% margin if s_x comes out 0.9. What is the smallest n that meets the target with t^*?
Using the interval from Exercise 2, is 4.0 pounds a plausible mean fleece weight for the yearlings? Is 4.2?
A neighbor claims Tupac’s llamas take 4.0 minutes each on average to shear. Using the data of Exercise 3, is 4.0 plausible at 99\% confidence? Construct the 95\% interval and answer again.
A small herd has 250 llamas. Find the sample size for a 95\% margin of at most 0.2 pound with \sigma = 0.9. What share of the herd is that, which condition does it strain, and which way is the usual interval off?

Reasoning

Explain why the interval uses t^* while the sample-size formula uses z^*, and why the sample size it gives is a slight underestimate.
Illari writes that “95\% of the herd’s fleeces weigh between 3.711 and 4.689 pounds.” Explain what is wrong and give a correct statement.
Explain why sampling four times as many llamas roughly halves the margin of error, and why with t^* the margin falls by slightly more than half.
Explain why one 9.0 pound fleece matters a great deal in a sample of 10 and much less in a sample of 60.

Answers

\mu is the mean fleece weight of all llamas in Tupac’s herd this year. Random; 10 \le 90; the values spread evenly from 3.1 to 5.6 with no outliers or strong skew. \bar{x} = 43.7 / 10 = 4.37, s_x \approx 0.7660, \text{SE} \approx 0.2422, \text{df} = 9, t^* = 2.262, margin \approx 0.548, interval 3.822 to 4.918. We are 95\% confident that the herd’s mean fleece weight is between 3.822 and 4.918 pounds.
Random; 20 \le 40; roughly symmetric. \displaystyle \text{SE} = 0.72 / \sqrt{20} \approx 0.1610, \text{df} = 19, t^* = 1.729, margin \approx 0.278, interval 3.572 to 4.128. We are 90\% confident that the mean first fleece of all the ranch’s yearlings is between 3.572 and 4.128 pounds.
\displaystyle \text{SE} = 1.1 / \sqrt{25} = 0.22, \text{df} = 24, t^* = 2.797, margin \approx 0.615, interval 3.985 to 5.215 minutes.
Margin \displaystyle \frac{4.7697 - 4.1303}{2} = 0.3197; \displaystyle \text{SE} = 0.6 / \sqrt{16} = 0.15; t^* = 0.3197 / 0.15 \approx 2.131, the 95\% value for 15 degrees of freedom, so the level is 95\%.
(a) No: a sample of 8 with an outlier (9.0 is far from the rest), so the normal condition fails. (b) Yes: random, 45 \le 90, and n \ge 30 with mild skew. (c) No: not a random sample; the easy-to-catch llamas may differ. (d) No: 90 is more than 60, a tenth of the herd.
\displaystyle \text{SE} = 0.8 / \sqrt{6} \approx 0.3266. Hers: 4.0 \pm 0.640, from 3.360 to 4.640. Correct: \text{df} = 5, t^* = 2.571, margin \approx 0.840, from 3.160 to 4.840. The t^* interval is right because s_x stands in for \sigma; her method captures \mu in only about 89\% of samples of 6.
1.729, for a 90\% interval with \text{df} = 19, so n = 20; 2.797, for a 99\% interval with \text{df} = 24, so n = 25.
\displaystyle n \ge \left(\frac{1.645 \times 0.9}{0.2}\right)^2 = 7.4025^2 \approx 54.80, so 55.
\displaystyle n \ge \left(\frac{1.96 \times 0.8}{0.1}\right)^2 = 15.68^2 \approx 245.86, so 246; a tenth of 3{,}000 is 300, and 246 \le 300.
\displaystyle \sigma = \frac{6 - 2}{4} = 1; \displaystyle n \ge \left(\frac{1.96}{0.3}\right)^2 \approx 42.68, so 43.
Predicted: \displaystyle 1.96 \times 0.9 / \sqrt{30} \approx 0.322 pound. With \text{df} = 29, t^* = 2.045: \displaystyle 2.045 \times 0.9 / \sqrt{30} \approx 0.336 pound.
With n = 55, \text{df} = 54, t^* = 1.674: margin \approx 0.203, just over 0.2. With n = 56 the margin is about 0.2012, still over; with n = 57 it is about 0.1994, so 57.
The interval is 3.572 to 4.128. 4.0 lies inside, so it is plausible; 4.2 lies above 4.128, so at 90\% confidence it is not.
At 99\% the interval is 3.985 to 5.215, which contains 4.0: plausible. At 95\%, t^* = 2.064, margin \approx 0.454, interval 4.146 to 5.054, which does not contain 4.0: evidence against the claim at that level. A narrower interval rules out more values.
\displaystyle n \ge \left(\frac{1.96 \times 0.9}{0.2}\right)^2 = 8.82^2 \approx 77.79, so 78, which is \displaystyle \frac{78}{250} = 0.312, about 31\% of the herd. That strains the tenth-of-the-population condition (a tenth is 25). Sampling so large a share without replacement makes \bar{x} vary less than the formula assumes, so the usual interval is wider than needed.
In the interval s_x estimates \sigma, and the standardized mean then follows a t curve with n - 1 degrees of freedom, so t^* gives the stated capture rate. In planning, n is unknown, so the degrees of freedom and t^* are too; z^* breaks the circle. Since every t^* is larger than the matching z^*, the real margin at the planned n comes out a little above m even when s_x equals the planning value, so the formula’s n is slightly too small.
The interval estimates the herd’s mean \mu, not where single fleeces fall; individual fleeces vary with standard deviation near 0.88 pound, far more than the mean does, so much less than 95\% of fleeces lie in that range. Correct: “We are 95\% confident that the mean fleece weight of all 600 llamas is between 3.711 and 4.689 pounds,” meaning the method captures \mu in about 95\% of random samples.
The margin is \displaystyle t^* s_x / \sqrt{n}, and \displaystyle \sqrt{4n} = 2\sqrt{n}, so the standard error halves. With t^*, the larger sample also has more degrees of freedom, so t^* itself is a little smaller, and the margin falls by slightly more than half.
In a sample of 10, one value far above the rest pulls \bar{x} up by a tenth of its distance from the others and inflates s_x a lot, so both the center and the width of the interval are driven by one animal, and the shape needed for t^* is not believable. In a sample of 60 its pull on \bar{x} is only a sixtieth of that distance, a sixth as large, and its effect on s_x smaller, and with n \ge 30 the sampling distribution of \bar{x} is close to normal anyway, so one such value matters much less (though it is still worth checking).