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
x̄ = 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.]
Suppose that BC financial aid alots a textbook stipend by claiming that the average textbook at...
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Do you want to test this claim. The Noel and alternative hypothesis
in symbols would be
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Test the claim that the proportion of people who own cats is significantly different than 40% at the 0.1 significance level. The null and alternative hypothesis would be: OH:P < 0.4 Hp > 0.4 OH = 0.4 H +0.4 H:p = 0.4 Hp + 0.4 Ho:p = 0.4 HP < 0.4 Hou < 0.4 H:H > 0.4 H: > 0.4 HAM < 0.4 The test is: two-tailed right-tailed left-tailed The test 15! two-tailed right-tailed left-tailed Based on a sample of...
Help with the following problem urget, bellow is the csv
data.
Campus
Online
153.83
148.66
212.21
214.58
191.65
191.24
142.49
141.73
154.71
152.62
223.36
224.79
93.93
92.87
112.21
106.41
182.58
185.15
179.55
177.33
187.51
185.17
168.75
167.08
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