Question

Suppose that BC financial aid alots a textboolk stipend by claiming that the average textbook at BC bookstore costs $ 94.98. You want to test this claim The null and alternative hypothesis in symbols would be: OHo:p 2 94.98 Hi:p94.98 Оно: р-94.98 Hi:p94.98 OHo H 94.98 H1 : μ > 94.98 Ho:p 94.98 Hi:p> 94.98 H0 : μ 94.98 H 94.98 0H0: μ > 94.98 H1: 94.98
The null hypothesis in words would be: OThe proportion of all textbooks from the store that are less than 94.98 is equal to 50% ○The average price of textbooks in a sample is OThe average of price of all textbooks from ○The average of price of all textbooks from 0 The average price of all textbooks from the $94.98 the store is greater than S 94.98. the store is less than $ 94.98 store is $ 94.98 Based on a sample of 190 textbooks at the store, you find an average of 95.78 and a standard deviation of 28.2 The point estimate is: decimals) (to 3 The 90 % confidence interval (use z*) is: (to 3 to decimals)
Based on this we OFail to reject the null hypothesis OReject the null hypothesis
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Answer #1

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pop. mean = 94.98

Ho: Mu = 94.98
Ha: Mu != 94.98

The null hypothesis in words would be:
4th option is correct:The average price of all textbooks from the store is $94.98

Also given is :

n = 190
Sample mean = 95.78
Stdev =28.2

Point estimate= sample mean ( because the sample mean is a unbiased estimator of pop. mean ) = 95.78

The 90% Confidence interval is : 1.645

So, 90% CI is 94.98 +/- 1.645*28.2/sqrt(190)

= 91.615 to 98.345

Since the sample mean (of 95.78) lies between the 90% Confidence interval ( 91.615, 98.345) we conclude that we fail to reject the null hypothesis

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