Answer to Question #342564 in Statistics and Probability for nana

Question

(a) Identify the claim and state H0 and Ha. (b) Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed, and whether to use a z-test, a t-test, or a chi-square test. Explain your reasoning. (c) Choose one of the options. If convenient, use technology. Option 1: Find the critical value(s), identify the rejection region(s), and find the appropriate standardized test statistic. Option 2: Find the appropriate standardized test statistic and the P-value. (d) Decide whether to reject or fail to reject the null hypothesis. (e) Interpret the decision in the context of the original claim.

7. A researcher claims that the mean age of the residents of a small town is more than 32 years. The ages (in years) of a random sample of 36 residents are listed below. At a = 0.10, is there enough evidence to support the researcher’s claim? Assume the population standard deviation is 9 years.

41 33 47 31 26 39 19 25 23 31 39 36

41 28 33 41 44 40 30 29 46 42 53 21

29 43 46 39 35 33 42 35 43 35 24 21

Answer 1

$$\bar{X}=\dfrac{1}{36}(41+33+47+31+26+39$$

$$+19+25+23+31+39+36+41+ 28$$

$$+ 33+ 41+ 44 +40+ 30+ 29+ 46+ 42$$

$$+ 53+ 21+ 29+ 43 +46 +39 +35 +33$$

$$+ 42+ 35+ 43 +35+ 24+ 21)=\dfrac{421}{12}\approx35.0833$$

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

$H_0:\mu\le32$

$H_1:\mu>32$

(b) corresponds to a right-tailed one-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

(c)Based on the information provided, the significance level is $\alpha = 0.10,$ and the critical value for a right-tailed test is $z_c =1.2816.$

The rejection region for this right-tailed test is $R = \{z:z>1.2816\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{\dfrac{421}{12}-32}{9/\sqrt{36}}\approx2.0556$$

(d) it is observed that $z=2.0556>1.2816=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(z>2.0556)=0.019911,$ and since $p= 0.019911<0.10=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is greater than 32, at the $\alpha = 0.10$ significance level.