Question

T-Test and the 5 steps.

A food Processing company is concerned that their sexteen ounce cans of sliced pineapple are being overilled. The quality control department took a random sample of fifty cans and found that he mean weight was 16.05 ounces, with a standard deviation of .03 ounces. Use the ninety-five percent confidence coeffiencent and test to determine wheather the company's concern is valid. Round allcalculations to four deimal places.

1) Null Hypthesis

2) Level of Significane

3) Decison rule

4) Math

5)DACR

Answer 1

Solution:-

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

Null hypothesis: u < 16.05
Alternative hypothesis: u > 16.00

Note that these hypotheses constitute a one-tailed test.  

Formulate an analysis plan. For this analysis, the significance level is 0.05. 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 = 0.00424
DF = n - 1

D.F = 49
t = (x - u) / SE

t = 11.79

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.

The observed sample mean produced a t statistic test statistic of 11.79.

Thus the P-value in this analysis is less than 0.0001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the company's concern is valid.