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raquelwilliams5795
27.02.2020 •
Mathematics
The owner of a local phone store wanted to determine how much customers are willing to spend on the purchase of a new phone. In a random sample of 14 phones purchased that day, the sample mean was $492.678 and the standard deviation was $26.4871. Calculate a 99% confidence interval to estimate the average price customers are willing to pay per phone.
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Ответ:
99% confidence interval bto estimate the average price customers are willing to pay per phone is between a lower limit of $471.356 and an upper limit of $514.
Step-by-step explanation:
Confidence interval = mean + or - Error margin (E)
mean = $492.678
sd = $26.4871
n = 14
degree of freedom = n - 1 = 14 - 1 = 13
confidence level = 99%
t-value corresponding to 13 degrees of freedom and 99% confidence level is 3.012
E = t×sd/√n = 3.012 × $26.4871/√14 = $21.322
Lower limit = mean - E = $492.678 - $21.322 = $471.356
Upper limit = mean + E = $492.678 + $21.322 = $514
99% confidence interval to estimate the average price customers are willing to pay per phone is between $471.356 and $514.
Ответ:
Step-by-step explanation:
7.008 is the answer