Question

(1 point) A sample of 6 measurments, randomly selected from a normally distributed population, resulted in a sample mean, t = 7.7 and sample standard deviation s = 1.2. Using a = 0.05, test the null hypothesis that the mean of the population is 7.2 against the alternative hypothesis that the mean of the population, j < 7.2 by giving the following: (a) the degree of freedom (b) the critical t value (c) the test statistic The final conclustion is A. We can reject the null hypothesis that u = 7.2 and accept that u < 7.2. B. There is not sufficient evidence to reject the null hypothesis that u = 7.2.

Answer 1

Solution :

Given that,

Population mean = $\mu$ = 7.2

Sample mean = $\bar x$ = 7.7

Sample standard deviation = s = 1.2

Sample size = n = 6

Level of significance = $\alpha$ = 0.05

This is a left tailed test.

Degrees of freedom = n - 1 = 5

Critical value of  the significance level is α = 0.05, and the critical value for a left-tailed test is

$t_{c}$ = -2.015

The test statistics,

t = ( $\bar x$ - $\mu$ )/ (s/$\sqrt{n}$)

= ( 7.7 - 7.2 ) / ( 1.2 /$\sqrt{}$6 )

=1.021

Since it is observed that t = 1.021 > $t_{c}$ = -2.015, it is then concluded that the null hypothesis is rejected.

Conclusion:

A) We can reject the null hypothesis that $\mu$ = 7.2 and accept that $\mu$ < 7.2