Question

Consider the following hypothesis test: H0: μ = 18 Ha: μ ≠ 18 A sample of 48 provided a sample mean x = 17 and a sample standard deviation s = 4.9. a. Compute the value of the test statistic (to three decimal places.) b. Use the t distribution table (Table 2 in Appendix B) to compute a range for the p-value. (to two decimal places) p-value is between is c. At α = .05, what is your conclusion? p-value is Selectgreater than or equal to 0.05, rejectgreater than 0.05, do not rejectless than or equal to 0.05, do not rejectless than 0.05, rejectequal to 0.05, do not rejectnot equal to 0.05, do not rejectItem 4 H0 d. What is the rejection rule using the critical value? Reject H0 if t is Selectgreater than or equal to -2.012greater than 2.012less than or equal to -2.012less than -2.012equal to 2.012not equal to -2.012Item 5 or t is Selectgreater than or equal to 2.012greater than -2.012less than or equal to 2.012less than -2.012equal to 2.012not equal to -2.012Item 6 What is your conclusion? t = ; Selectdo not rejectrejectItem 8 H0

Answer 1

Here

sample mean $\bar{x} = 17$

sample standard deviation

and sample size

The test statistic can be written as

$t = \frac{\sqrt{n}(\bar{x} - 18)}{s}$ which under H₀ follows a t distribution with n-1 df.

We reject H₀ at 5% level of significance if or P-value < 0.05

Now,

The value of the test statistic

and critical value $t_{critical} = \pm t_{0.025,n-1} = \pm t_{0.025,47} = \pm 2.012$

and P-value

Since P-value > 0.05 and $|t_{obs}| = 1.41392 \ngtr 2.012$ , so we fail to reject H₀ at 5% level of significance and we can conclude that the population mean is not significantly different from 18.