Question

According to a recent report, the mean number of meals eaten at home is less than 700 per year. Suppose that a random sample of 100 households showed a sample mean number of meals prepared at home of 650. Assume the population standard deviation is 25. Test the claim that the population mean is less than 700 using a significance level of 0.05. What is the null hypothesis? What is the alternate hypothesis? Based on the problem statement, what is α Which test will you use to test the null hypothesis? Please explain why you will use this test. What is your calculated p-value? How does the p-value compare with α ? What is your conclusion?

Answer 1

Answer:

We conclude that the population mean is less than 700 per year.

Step-by-step explanation:

We are given that the mean number of meals eaten at home is less than 700 per year. Suppose that a random sample of 100 households showed a sample mean number of meals prepared at home of 650.

Assume the population standard deviation is 25.

Let  = mean weekly earnings of a production worker.

So, Null Hypothesis, $H_0$ : $\mu$ = 700     {means that the population mean is equal to 700 per year}

Alternate Hypothesis, $H_A$ : $\mu$ < 700     {means that the population mean is less than 700 per year}

The test statistics that would be used here One-sample z test statistics because 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 number of meals prepared at home = 650

            $\sigma$ = population standard deviation = 25

            n = sample of households = 100

So, the test statistics  =  $\frac{650-700}{\frac{25}{\sqrt{100} } }$

                                     =  -20

The value of z test statistics is -20.

Also, P-value of the test statistics is given by;

          P-value = P(Z < -20) = 1 - P(Z $\leq$ 20)

                        = 1 - 0.9999 = 0.0001

Here also, p-value is less than the level of significance, so we reject our null hypothesis.

Here, $\alpha$ (level of significance) is given as 0.05 or 5%. Now, at 5% significance level the z table gives critical value of -1.645 for left-tailed test.

Since our test statistic is less than the critical value of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the population mean is less than 700 per year.