Answer to Question #217480 in Statistics and Probability for Shipra

Question

The heights of ten children selected at random from a given locality had a mean 63.2cm and variance 6.25cm. Test at 5% level of significance, the hypothesis that the children of the given locality are on average less than 65 cm. given for 9 degrees of freedom P(t>1.83)=0.05

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq65$

$H_1:\mu<65$

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha=0.05,$ $df=n-1=10-1=9$ degrees of freedom, and the critical value for a left-tailed test is $t_c=-1.833113.$

The rejection region for this left-tailed test is $R=\{t:t<-1.833113\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{63.2-65}{6.25/\sqrt{10}}=-0.910736$$

Since it is observed that $t=-0.910736>-1.833113=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value for left-tailed, $\alpha=0.05, df=9,$ $t=-0.940736$ is $p=0.193088,$ and since $p=0.193088>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is less than $65,$ at the $\alpha=0.05$ significance level.