Question

1. A researcher estimates that the average height of the buildings of 30 or more stories in a large city is at least 700 feet. A random sample of 10 buildings is selected, and the heights in feet are shown. At  = 0.025, is there enough evidence to reject the claim? 485 511 841 725 615 520 535 635 616 582

Answer 1

$$\bar{x}=\dfrac{1}{10}(485 +511+ 841+ 725+ 615$$

$$+520 +535 +635 +616+ 582)=606.5$$

$$s^2=\dfrac{\sum_i(x_i-\bar{x})^2}{n-1}=\dfrac{1}{10-1}((485-606.5)^2$$

$$+(511-606.5)^2+(841-606.5)^2$$

$$+(725-606.5)^2+(615-606.5)^2$$

$$+(520-606.5)^2+(535-606.5)^2$$

$$+(635-606.5)^2+(616-606.5)^2$$

$$+(582-606.5)^2)=\dfrac{107084.5}{9}$$

$$s=\sqrt{s^2}=\sqrt{\dfrac{107084.5}{9}}\approx109.08$$

The parameter is average content of fruit concentrate per bottle.

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge700$

$H_1:\mu<700$

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 $\alpha = 0.025,$ $df=n-1=9$ and the critical value for a left-tailed test is $t_c =-2.262156.$

The rejection region for this left-tailed test is $R = \{t:t<-2.262156\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{606.5-700}{109.08/\sqrt{10}}=-2.7106$$

Since it is observed that $t=-2.7106<-2.262156=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for left-tailed, $df=9$ degrees of freedom, $t=-2.7106$ is $p=0.011987,$ and since $p=0.011987<0.025=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is less than 700, at the $\alpha = 0.025$ significance level.