Question

A local bottler in Hawaii wishes to ensure that an average of 22 ounces of passion fruit juice is used to fill each bottle. In order to analyze the accuracy of the bottling process, he takes a random sample of 65 bottles. The mean weight of the passion fruit juice in the sample is 21.54 ounces. Assume that the population standard deviation is 1.38 ounce.
a. Select the null and the alternative hypotheses for the test.

1. H0: μ = 22; HA: μ ≠ 22
2. H0: μ ≤ 22; HA: μ > 22
3. H0: μ ≥ 22; HA: μ < 22

b. Calculate the value of the test statistic.
c. Find the critical value(s).

d. What is the conclusion?
1. Reject H0 since the value of the test statistic is not less than the negative critical value.
2. Reject H0 since the value of the test statistic is less than the negative critical value.
3. Do not reject H0 since the value of the test statistic is not less than the negative critical value.
4. Do not reject H0 since the value of the test statistic is less than the negative critical value.

Answer 1

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = 22 ounces  

    Alternate Hypothesis, $H_A$ : $\mu\neq$ 22 ounces

(b) The value of the test statistic is -2.687.

(c) The critical values are -1.96 and 1.96.

(d) We conclude that Reject $H_0$ since the value of the test statistic is less than the negative critical value.

Step-by-step explanation:

We are given that a local bottler in Hawaii wishes to ensure that an average of 22 ounces of passion fruit juice is used to fill each bottle.

He takes a random sample of 65 bottles. The mean weight of the passion fruit juice in the sample is 21.54 ounces. Assume that the population standard deviation is 1.38 ounce.

Let $\mu$ = average ounces of passion fruit juice used to fill each bottle.

(a) So, Null Hypothesis, $H_0$ : $\mu$ = 22 ounces     {means that the average of 22 ounces of passion fruit juice is used to fill each bottle}

Alternate Hypothesis, $H_A$ : $\mu\neq$ 22 ounces     {means that the average different from 22 ounces of passion fruit juice is used to fill each bottle}

The test statistics that would be used here 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 weight of the passion fruit juice = 21.54 ounces

            $\sigma$ = population standard deviation = 1.38 ounce

            n = sample of bottles = 65

So, test statistics  =  $\frac{21.54-22}{\frac{1.38}{\sqrt{65} } }$  

                               =  -2.687

(b) The value of z test statistics is -2.687.

(c) Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics does not lie within the range of critical values 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 Reject $H_0$ since the value of the test statistic is less than the negative critical value which means that the average different from 22 ounces of passion fruit juice is used to fill each bottle.