Question

Previous Question Question 14 of 20 Next Question A company randomly surveys 15 VIP customers and records their customer satisfaction scores out of a possible 100 points. Based on the data provided, calculate a 90% confidence interval to estimate the true satisfaction score of all VIP customers.

Answer 1

71.4699, 81.7301.

Explanation:

So, the following are the scores: 74

90

84

78

61

65

62

67

73

75

76

95

71

98

80

`Let,\;the\;total\;sum\;of\;the\;scores\;be:\sum x =1149`

`Then,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum x^2=89795`

The length of the score : 15

Then, let the size of the scores be n : 15.

Then, we have to find the mean.

`\bar{x}=(\sum x)/(n)`

`=(1149)/(15)=76.6`

`Then,\;we\;have\;to\;find\;standard\;deviation:S=\sqrt{\frac{\sum x^2-n.(\bar{x})^2}{n-1} }`

`=\sqrt{(89795-15*(76.6)^2)/(15-1) }=11.2808`

`The\;degree\;of\;freedom:n-1=14`

`So,\;the\;significant\;level:\alpha=1-0.90=0.10`

`and\;the\;critical\;value:t_{(\alpha)/(2) }=1.7613`

`Finally:\\=\bar{x}\;\pm\;t_{(\alpha)/(2) }\;(S)/(√(n) )\\=(71.4699, 81.7301)`