Question

Peanut:A company that packages salted peanuts in 8-oz. jars is interested in maintaining control on the amount of peanuts put in jars by one of its machines. Control is defined as averaging 8 oz. per jar and not consistently over-or under-filling the jars. To monitor this control, 16 jars were picked from the line at random time intervals and their contents weighed. The mean weight of peanuts in these 16 jars was 7.89 oz with a standard deviation of 0.2 oz. Based on this sample data; test the hypothesis at 5 percent significance level that the machine is indeed working properly. If it is deemed not to be doing so, recommend the management to conduct a costly adjustment.

1. State the null and alternate hypotheses
2. Identify critical values, and calculate test statistics
3. Draw conclusion and make recommendation to the management whether machine requires costly adjustment or not.
4. Estimate or calculate p-value

Answer 1

Solution:

The null and alternative hypotheses are as follows:

H_{​​​​​​0} : μ = 8 oz i.e. The population mean of the weight of peanuts in the jars is 8 oz.

H_{​​​​​1} : μ ≠ 8 oz i.e. The population mean of the weight of peanuts in the jars is not equal to 8 oz.

To test hypothesis we shall use one sample t-test. The test statistic is given as follows:

Where, x̄ is sample mean, μ is hypothesized value of population mean, s is sample standard deviation and n is sample size.

We have, x̄ = 7.89 oz, μ = 8 oz, s = 0.2 oz and n = 16

The value of the test statistic is -2.2

Since, our test is two-tailed test, therefore we shall obtain two-tailed critical-value for the t-test. At significance level of 5% = 0.05 and n-1 = 16-1 = 15 degrees of freedom the two-tailed critical-value for the t-test is given as follows:

Critical value = t_{(0.05/2, 15)} = 2.1314

For two-tailed t-test we make decision rule as follows:

If |t| > t_{(0.05/2, 15)} then we reject the null hypothesis (H_{​​​​​​0}) at 0.05 significance level.

If |t| < t_{(0.05/2, 15)} then we fail to reject the null hypothesis (H_{​​​​​​0}) at 0.05 significance level.

We have, |t| = 2.2 and t_{(0.05/2, 15)} = 2.1314

(2.2 > 2.1314)

Since, |t| > t_{(0.05, 15)}, therefore we reject the null hypothesis (H_{​​​​​​0}) at 0.05 significance level.

Conclusion : At 5% significance level there is not sufficient evidence to support that the machine is working properly.

Hence, I recommend the management to conduct a costly adjustment.

The two-tailed p-value is given as follows:

p-value = 2.P(T > |t|)

We have, |t| = 2.2

Hence, p-value = 2.P(T > 2.2)

p-value = 0.0439

The p-value is 0.0439.

Please rate the answer. Thank you.