Question

A machine makes twist-off caps for bottles. The machine is adjusted to make caps of diameter 1.85. Production records show that when the machine is so adjusted it will make caps with a mean diameter of 1.85 cm and with a standard deviation of 0.05 cm. During production, an inspector checks the diameters of caps to see if the machine has slipped out of adjustment. A random sample of 64 caps is taken. If the mean diameter for the sample is 1.87, does this indicate that the machine has slipped out of adjustment and the average diameter is different than 1.85? (use a 1% level of significance).

*Please show the process of solving the answer, specifically in Excel*

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 1.85
Alternative hypothesis: u 1.85

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.00078125

DF = n - 1

D.F = 63
t = (x - u) / SE

t = 3.20

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesised population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the t statistic having 63 degrees of freedom is less than -3.20 or greater than 3.20.

P-value = P(t < - 3.20) + P(t > 3.20)

Use the calculator to determine the p-values.

P-value = 0.001 + 0.001

Thus, the P-value = 0.00215

Interpret results. Since the P-value (0.002) is less than the significance level (0.01), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the average diameter is different than 1.85.