Question

4. A food company is concerned that its 16-ounce can of sliced pears is being overfilled. The quality-control department took a sample of 35 cans and found that the sample mean weight was 16.04 ounces, with a sample standard deviation of 0.08 ounces. Test at a 5% level of significance to see if the mean weight is greater than 16 ounces. What is your conclusion?

Answer 1

The mean weight is not greater than 16 ounces

Explanation:

Null hypothesis: Mean weight is greater than 16 ounces

Alternate hypothesis: Mean weight is not greater than 16 ounces.

Sample mean= 16.04 ounces

Assumed mean= 16 ounces

Significance level = 0.05

Sample standard deviation = 0.08 ounces

Number of samples= 35

To test the claim about the mean, Z = (sample mean - assumed mean) ÷ (standard deviation÷ √number of samples)

Z = (16.04 - 16) ÷ (0.08 ÷ √35)

Z = 0.04 ÷ (0.08 ÷ 5.92)

Z= 0.04 ÷ 0.014 = 2.86

Z = 2.86

It is a one-tailed test, the critical region is an area of 0.05 (the significance level). The critical value of the critical region is 1.64

Since the computed Z 2.86 is greater than the critical value 1.64, the null hypothesis is rejected.

Conclusion

The mean weight is not greater than 16 ounces