1 answer

Suppose that BC financial aid alots a textbook stipend by claiming that the average textbook at...

Question:

Suppose that BC financial aid alots a textbook stipend by claiming that the average textbook at BC bookstore costs $62.74. Yo

Suppose that BC financial aid alots a textbook stipend by claiming that the average textbook at BC bookstore costs $62.74. You want to test this claim. The null and alternative hypothesis in symbols would be: OH :p=62.74 Hp+62.74 O Hou < 62.74 HM > 62.74 OHO:n = 62.74 H: 62.74 OH: > 62.74 H:H<62.74 OH :p < 62.74 Hp > 62.74 OH,:p> 62.74 The null hypothesis in words would be: O The average price of textbooks in a sample is $ 62.74 O The average of price of all textbooks from the store is greater than $ 62.74. The proportion of all textbooks from the store that are less than 62.74 is equal to 50% The average of price of all textbooks from the store is less than $ 62.74. O The average price of all textbooks from the store is $ 62.74 Based on a sample of 110 textbooks at the store, you find an average of 73.74 and a standard deviation of 10. The point estimate is: (to 3 decimals) to (to 3 The 95% confidence interval (use 2*) is: decimals) Based on this we: O Fail to reject the null hypothesis O Reject the null hypothesis

Answers

Solution:

The null and alternative hypothesis in symbol would be:

H0: µ = 62.74

H1: µ ≠ 62.74

The null hypothesis in words would be:

The average price of all textbooks from the store is $62.74.

The point estimate is 73.740.

[The point estimate for population mean is the sample mean.]

Confidence interval for Population mean is given as below:

Confidence interval = x̄ ± Z*σ/sqrt(n)

From given data, we have

= 73.74

σ = 10

n = 110

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence interval = x̄ ± Z*σ/sqrt(n)

Confidence interval = 73.74± 1.96*10/sqrt(110)

Confidence interval = 73.74 ± 1.96*0.9535

Confidence interval = 73.74 ± 1.8688

Lower limit = 73.74 - 1.8688 = 71.871

Upper limit = 73.74 + 1.8688 =75.609

Confidence interval = (71.871, 75.609)

The 95% confidence interval is: 71.871 to 75.609

Based on this we:

Reject the null hypothesis

[We reject the null hypothesis because the hypothesized value of µ = 62.74 is not lies within the above confidence interval.]

.

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