Mathematics: A random sample of 10 scores on a recent chemistry exam is given below: 83818587797384867884 You may take as known that the population is normal with a standard deviation of 13. Carry out a hypothesis test (filling in the requested information) to determine if there is enough evidence at a 0.005 level of significance to infer that the mean score for this population exceeds 73.

A random sample of 10 scores on a recent chemistry

Ответ:

harlon852

22.01.2022

Yes, we have enough evidence at a 0.005 level of significance to infer that the mean score for this population exceeds 73.

Step-by-step explanation:

We are given that a random sample of 10 scores on a recent chemistry exam is given below:

83, 81, 85, 87, 79, 73, 84, 86, 78, 84

You may take as known that the population is normal with a standard deviation of 13.

We have to conduct a hypothesis test to infer that the mean score for this population exceeds 73.

Let $\mu$ = mean score for this population

SO, Null Hypothesis, H_0 : $\mu \leq$ 73 {means that the mean score for this population is less than or equal to 73}

Alternate Hypothesis, H_a : $\mu$ > 73 {means that the mean score for this population exceeds 73}

The test statistics that will be used here is One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean score = $\frac{83+81+85+87+79+73+84+86+78+84}{10}$ = $\frac{820}{10}$ = 82

$\sigma$ = population standard deviation = 13

n = sample size = 10

So, test statistics = $\frac{82-73}{\frac{13}{\sqrt{10} } }$

= 2.189

So, at 0.005 level of significance, the z table gives critical value of 3.8906 for one-tailed test. Since our test statistics is less than the critical value of z so we have insufficient evidence to reject null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the mean score for this population exceeds 73.