Solution :
=> from the given information,
Null and Alternative hypothesis :
H0 : μ = 80.87
H1 : μ ≠ 80.87
=============================================
Solution :
Given that n = 11, mean x-bar = 33 , standard deviation s = 2
=> df = n - 1 = 10
=> for 80% confidence level, t = 1.372
=> A 80% confidence interval of the mean is
=> x-bar +/- t*s/sqrt(n)
=> 33 +/- 1.372*2/sqrt(11)
=> 33 +/- 0.8273
=> 33.0 +/- 0.8 (rounded)
Suppose that easy financial aid a lot of text book stipend by claiming that the average...
1. Suppose that BC financial aid alots a textbook stipend by claiming that the average textbook at BC bookstore costs $$ 67.29. You want to test this claim. Based on a sample of 130 textbooks at the store, you find an average of 66.69 and a standard deviation of 18.4. The point estimate is: ___ (to 3 decimals) The 95 % confidence interval (use z*) is: ___to ___ (to 3 decimals) 2. Test the claim that the proportion of people...
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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...
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In a survey, 11 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $33 and standard deviation of $2. Construct a confidence interval at a 80% confidence level. Give your answers to one decimal place. + I Question Help: Message instructor Submit Question O e ORI Type here to search C . 5 6 8
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