Question

You wish to test the following claim ( H 1 ) at a significance level of α = 0.005 . H 0 : μ = 63.4 H 1 : μ ≠ 63.4 You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n = 7 with mean ¯ x = 39.3 and a standard deviation of s = 15.2 .

What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic =

What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value =

Answer 1

Solution:-

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

Null hypothesis: u = 63.4
Alternative hypothesis: u 63.4

Note that these hypotheses constitute a two-tailed test.

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 = 5.7451

DF = n - 1

D.F = 6
t = (x - u) / SE

t = - 4.195

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

Since we have a two-tailed test, the P-value is the probability that the t statistic having 6 degrees of freedom is less than -4.195 or greater than 4.195.

Thus, the P-value = 0.0057

Interpret results. Since the P-value (0.0057) is greater than the significance level (0.005), we cannot reject the null hypothesis.