Answer to Question #272360 in Statistics and Probability for Hunter

Question

The data below gives the end year profit in million shillings of twelve randomly selected pharmaceutical firms in Nairobi county;71.2,52.7,95.7,63.5,91.8,102.7,87.6,77.9,81.8,90.1,123.7,79.9.It is known that the average end of year profit of the twelve firms is 89.91m. An average value less than 89.91m is considered unacceptable.

a.)Test the null hypothesis at 5% level of significance. What is the value of your standard deviation, tabulated and calculated test static and decision?

b.) Construct a 95% confidence interval for the sample mean

Answer 1

$$mean=\bar{x}=\dfrac{1}{12}\sum_ix_i=\dfrac{1017.7}{12}$$

$$\approx84.8083$$

$$s^2=\dfrac{1}{12-1}\sum _i(x_i-\bar{x})^2\approx345.151742$$

$$s=\sqrt{345.151742}\approx18.57826$$

a) The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq89.91$

$H_1;\mu<89.91$

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 $\alpha = 0.05,$ $df=12-1=11$ degrees of freedom, and the critical value for a left-tailed test is $t_c = -1.795885.$

The rejection region for this left-tailed test is $R = \{t: t < -1.795885\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{84.8083-89.91}{18.57826/\sqrt{12}}$$

$$=-0.951263$$

Since it is observed that $t = -0.951263 \ge -1.795885=t_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value for left-tailed, $\alpha=0.05, df=11$ degrees of freedom, $t = -0.951263$ is $p=0.180952,$ and since $p= 0.180952>0.05=\alpha,$ 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 $89.91,$ at the $\alpha = 0.05$ significance level.

b) The critical value for $\alpha = 0.05$ and $df = n-1 = 11$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.200985.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(84.8083- 2.200985\times\dfrac{18.57826}{\sqrt{12}},$$

$$84.8083+2.200985\times\dfrac{18.57826}{\sqrt{12}})$$

$$=(73.004, 96.612)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $73.004 < \mu < 96.612,$ which indicates that we are 95 % confident that the true population mean $\mu$ is contained by the interval $(73.004, 96.612).$