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?

The mean weight is not greater than 16 ounces

Step-by-step 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

you could deduce the conclusion directly using the facts pertaining to the specific situation.

hope this .