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?

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

“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).”

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