Question

The following two sample data sets both have sample mean 6. Set I 13.4 3.7 0.8 103 1.8 Set II 2.6 8.2 5.2 7.2 6.8 (a) If μ is the population mean perform t tests for each set to test Ho : ,,-10 against H 11 #10. (b) Do you consider the conclusions of the tests reasonable? Do you have any reservations about Use significance level a 0.05. using a t-test for either of these data sets?

Answer 1

Solution:-

a)

For Set I

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

Null hypothesis: u = 10
Alternative hypothesis: u $\neq$ 10

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it 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 = 2.4821

DF = n - 1

D.F = 4
t = (x - u) / SE

t = -1.61

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 4 degrees of freedom is less than -1.61 or greater than 1.61.

Thus, the P-value = 0.183.

Interpret results. Since the P-value (0.183) is greater than the significance level (0.05), we cannot reject the null hypothesis.

For Set II

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

Null hypothesis: u = 10
Alternative hypothesis: u $\neq$ 10

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it 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 = 1.0127

DF = n - 1

D.F = 4
t = (x - u) / SE

t = - 3.89

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 4 degrees of freedom is less than -3.89 or greater than 3.89.

Thus, the P-value = 0.018

Interpret results. Since the P-value (0.018) is less than the significance level (0.05), we have to  reject the null hypothesis.

b) Yes the conclusions of the tests are reasonable.