Suppose we obtain n = 100 identically and independently distributed draws of the

Question

Suppose we obtain n = 100 identically and independently distributed draws of the random variable Xi . The sample average is ˆµ = 51.6. We have an estimate of the standard deviation of Xi , ˆ = 1.3.

i. Suppose we specify the null hypothesis µ0 = 50, but we only want to reject when the sample average is above the value under the null hypothesis. At the = .05 signifiance level, conduct a hypothesis test that only rejects when the sample average is above the value under the null hypothesis.

ii. Suppose we learn that there is a data error. Because of this error, the 1st and 2nd observations were perfectly correlated and the same value, the 3rd and 4th observations were perfectly correlated and the same value, etc. However, the 1st, 3rd, 5th, etc. observations are each still independent and identically distributed.

iii.With this data error discovered, at the = .05 signifiance level, test the null hypothesis that the population mean of Xi is 50, against a general two-sided alternative hypothesis. (Hint: we can think of the original dataset as being comprised of two datasets of 50 independent and identically distributed observations.)

Explanation / Answer

Solution:-

i)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: < 50
Alternative hypothesis: > 50

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.13
DF = n - 1

D.F = 99
t = (x - ) / SE

t = 12.31

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of 12.31.

Thus the P-value in this analysis is less than 0.0001.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.

iii)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: < 50
Alternative hypothesis: > 50

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.18385
DF = n - 1

D.F = 49
t = (x - ) / SE

t = 8.70

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of 8.70

Thus the P-value in this analysis is less than 0.0001.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.

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