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16 July, 04:55

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?

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  1. 16 July, 05:43
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    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
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