Answer to Question #338719 in Statistics and Probability for Seb

Question

A fruit juice franchise company has a policy of opening new fruit juice stand only on those areas that have a mean household income of at least ₱30,500 a month. The company is currently considering an area in which to open a new fruit juice stand. The company’s research department took a sample of 25 households from this area and found that the mean monthly income of these households is ₱32,600. Using 5% significance level, would you conclude that the company should open a fruit juice stand in the area? Also, find the 95% confidence interval of the true mean.

Answer 1

Let standard deviation is $s=4500.$

We need to construct the $95\%$ confidence interval for the population mean $\mu.$

The critical value for $\alpha = 0.05$ and $df = n-1 = 24$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.063899.$

The corresponding confidence interval is computed as shown below:

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

$$=(32600-2.063899\times\dfrac{4500}{\sqrt{25}},$$

$$32600+2.063899\times\dfrac{4500}{\sqrt{25}})$$

$$=(30742.4909,34457.5091)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $30742.4909 < \mu < 34457.5091,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(30742.4909, 34457.5091).$

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge30500$

$H_a:\mu<30500$

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 $\alpha = 0.05,$ $df=n-1=24$ degrees of freedom, and the critical value for a left-tailed test is $t_c = -1.710882.$

The rejection region for this left-tailed test is $R = \{t: t < -1.710882\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{32600-30500}{4500/\sqrt{25}}\approx2.3333$$

Since it is observed that $t = 2.3333 \ge -1.711=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for left-tailed test, $df=24$ degrees of freedom, $t=2.3333$ is $p = 0.98583,$ and since $p = 0.98583 \ge 0.05=\alpha,$ 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 30500, at the $\alpha = 0.05$ significance level.