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

Answer:

71.4699, 81.7301.

Step-by-step 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.

                                      $\bar{x}=\frac{\sum x}{n}$

                                         $=\frac{1149}{15}=76.6$

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

                                         $=\sqrt{\frac{89795-15\times(76.6)^2}{15-1} }=11.2808$

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

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

$and\;the\;critical\;value:t_{\frac{\alpha}{2} }=1.7613$

$Finally:\\=\bar{x}\;\pm\;t_{\frac{\alpha}{2} }\;\frac{S}{\sqrt{n} }\\=(71.4699, 81.7301)$