Based on the information from the package, the average weight of a type of choco

Question

Based on the information from the package, the average weight of a type of chocolate pie should be 50 grams. If we randomly draw a sample of 15 this type of chocolate pies measure their weights. The average is 48.6 grams, with standard deviation 0.8 grams. (1) Is the average weight of the chocolate pie below 50 grams? Use significant level 0.05. (2) Is the average weight of this chocolate pie different from 50 grams? use significant level 0.05. Please explain where exactly and which table you used to get the t score because that confuses and offer explanations why it's different. Thank you so much for your help.

Explanation / Answer

1)
Hypothesis:
H0 : mu = 50
Ha : mu < 50

Data:
mean = 50 , x = 48.6 , s = 0.8 , ,n = 15
Test statistic:
t = ( x - mean) / (s/sqrt(n))
= ( 48.6 - 50) / ( 0.8/ sqrt(15))
= -6.77

P value is calculated using t = -6.77 , df = 14

p value = 0.00001. (to get this value use t-table, as this is one tailed test use left tailed table)

In order to get this p-value, you can use calculator. Or one can use this link to find p-value - http://www.socscistatistics.com/pvalues/tdistribution.aspx

As p-value is less than significance level, we reject the null hypothesis.

2)

Hypothesis:
H0 : mu = 50
Ha : mu not equal to 50

Data:
mean = 50 , x = 48.6 , s = 0.8 , ,n = 15
Test statistic:
t = ( x - mean) / (s/sqrt(n))
= ( 48.6 - 50) / ( 0.8/ sqrt(15))
= -6.77

P value is calculated using t = -6.77 , df = 14

p value = 0.00001. (As value of t is very left to the curve, p-value will not change significantly from the above test)
As p-value is less than significance level of 0.05, we reject the null hypothesis.

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