Success, a recently released magazine targeted to young professionals claims tha

Question

Success, a recently released magazine targeted to young professionals claims that 65% of its subscribers have an annual income greater than $120,000. The sales staff at Success uses this high proportion of subscribers earning more than $120,000 as a selling point when selling advertising. Success is currently trying to sell a large amount of advertising to a company that sells expensive wines. The wine company’s market research division decides to conduct a survey of 196 of Success subscribers to verify the claim that 65% of its subscribers earn more than $120,000 annually. The file Success contains the results of the survey. The responses are coded “Yes” is the subscriber earns more than $120,000 annually and “No” if the subscriber does not earn more than $120,000 annually. Is there evidence to support the magazine company’s claim at the 5% level.

A. (1 pt) State appropriate hypotheses.

B. (1 pt) State and verify the conditions for this test of significance.

C. (2 pts) Using Minitab, perform the appropriate test of significance. Provide Minitab output.

D. (2 pts) Based upon this test of significance, what can the wine company conclude about the subscribers of this magazine?

E. (1 pt) Describe Type 1 error in this situation and give a possible consequence of this error.

F. (1 pt) Describe Type 2 error in this situation and given a possible consequence of this error.

1 No

2 Yes

3 Yes

4 Yes

5 Yes

6 Yes

7 No

8 Yes

9 No

10 Yes

11 Yes

12 Yes

13 Yes

14 No

15 Yes

16 Yes

17 Yes

18 Yes

19 Yes

20 Yes

21 Yes

22 Yes

23 No

24 Yes

25 Yes

26 Yes

27 No

28 Yes

29 Yes

30 Yes

31 Yes

32 No

33 No

34 Yes

35 Yes

36 Yes

37 Yes

38 No

39 Yes

40 Yes

41 No

42 Yes

43 No

44 Yes

45 Yes

46 Yes

47 Yes

48 Yes

49 Yes

50 Yes

51 Yes

52 Yes

53 Yes

54 No

55 Yes

56 Yes

57 Yes

58 Yes

59 No

60 Yes

61 Yes

62 Yes

63 Yes

64 Yes

65 No

66 No

67 No

68 Yes

69 No

70 Yes

71 No

72 No

73 Yes

74 Yes

75 Yes

76 No

77 Yes

78 No

79 Yes

80 Yes

81 Yes

82 No

83 No

84 No

85 No

86 No

87 Yes

88 No

89 Yes

90 Yes

91 Yes

92 Yes

93 Yes

94 No

95 No

96 Yes

97 No

98 Yes

99 Yes

100 No

101 No

102 Yes

103 Yes

104 Yes

105 No

106 No

107 Yes

108 Yes

109 Yes

110 Yes

111 Yes

112 Yes

113 Yes

114 Yes

115 No

116 No

117 Yes

118 Yes

119 Yes

120 No

121 Yes

122 Yes

123 Yes

124 No

125 Yes

126 No

127 Yes

128 No

129 No

130 Yes

131 Yes

132 No

133 No

134 Yes

135 Yes

136 No

137 Yes

138 Yes

139 Yes

140 Yes

141 Yes

142 No

143 Yes

144 Yes

145 Yes

146 No

147 No

148 No

149 Yes

150 Yes

151 Yes

152 Yes

153 No

154 Yes

155 Yes

156 Yes

157 Yes

158 Yes

159 No

160 Yes

161 No

162 Yes

163 Yes

164 No

165 No

166 No

167 No

168 Yes

169 Yes

170 No

171 Yes

172 No

173 Yes

174 No

175 Yes

176 Yes

177 No

178 No

179 Yes

180 Yes

181 Yes

182 Yes

183 Yes

184 No

185 No

186 Yes

187 Yes

188 Yes

189 Yes

190 No

191 No

192 No

193 Yes

194 Yes

195 Yes

196 Yes

Explanation / Answer

Solution:-

Yes = 131

No = 65

p = 131/196

p = 0.6684

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P = 0.65
Alternative hypothesis: P 0.65

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample proportion is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

= sqrt[ P * ( 1 - P ) / n ]

= 0.0341
z = (p - P) /

z = 0.54

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 0.54 or greater than 0.54.

Thus, the P-value = 0.5892

Interpret results. Since the P-value (0.5892) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that 65% of its subscribers earn more than $120,000 annually.

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