Question

2) A cereal company claims that the mean weight of the cereal in its packets is 14 oz. The weights (in ounces)
of the cereal in a random sample of 8 of its cereal packets are listed below.
14.6 13.8 14.1 13.7 14.0 14.4 13.6 14.2
Test the claim at the 0.01 significance level.

using it ti-84 calculator please stpe by step

Answer 1

Solution:-

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

Null hypothesis: u = 14
Alternative hypothesis: u 14

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 = 0.12247

DF = n - 1

D.F = 7
t = (x - u) / SE

t = 0.41

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized 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 7 degrees of freedom is less than -0.41 or greater than 0.41.

Thus, the P-value = 0.694.

Interpret results. Since the P-value (0.694) is greater than the significance level (0.01), we cannot reject the null hypothesis.