Question

The following sample of eight measurements was randomly selected from a normally distributed population: 12, 10, 9, 8, 15, 10, 11, and 13. Test for significant difference between the sample mean and the population mean of 10. Use 𝛼 = 0.05.

Answer 1

First determine the value of the sample mean and standard deviation.

The mean is given by,

$\bar{x}=\sum(x)/n$ where $n=8$

$\bar{x}=88/8=11$

To find the standard deviation, we need to find the variance given by the formula,

$V(X)=(\sum(x^2)-(\sum(x))^2/n)/(n-1)$

$V(X)=(1004-968)/7=5.142857$

The standard deviation $SD(X)=\sqrt{V(X)}=\sqrt{5.142857}=2.267787$

The hypotheses tested are,

$H_0:\mu=10\space Against\space H_1:\mu\not=10$

To perform this test, we shall apply the t distribution since the population variance is not known and the sample size is less than 30.

The test statistic is given as,

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

$t=(11-10)/(2.267787/\sqrt{8})=1/0.801784=1.247219$

$t$ is compared with the table value at $\alpha=0.05$ with $(n-1)=8-1=7$ degrees of freedom.

Thus, $t_{\alpha/2,7}=t_{0.05/2,7}=t_{0.025,7}=2.365$, and the null hypothesis is rejected if $t\gt t_{0.025,7}$

Since $t=1.247219\lt t_{0.025,7}=2.365$, we fail to reject the null hypothesis and we conclude that there is no sufficient evidence to show that the sample mean and the population mean are significantly different at 5% significance level.