ck 1 1 145739999256 0034quiz action-takeQuiz&iquiz; probGuid-ONAPCOA801010000002

Question

ck 1 1 145739999256 0034quiz action-takeQuiz&iquiz; probGuid-ONAPCOA801010000002 Due Tuesday 03.08.16 at 220 Graded Assignment Attempts Average: 112 Aa Aa 5. Hypothesls tests about a population proportion A firm's corporate strategy is driven largely by its top management team. One method of gauging the marketing on corporate strategy is to measure the proportion of firms with a chief marketing officer on their top influence of management team. Over the 5-year period from 2000 to 2004, 42% of firms had a chief marketing officer on ther to management team. [ Source: Pravin Nath and Vijay Mahajan, "Chief Marketing Officers: A Study of Their Presence in Firms, Top Management Teams,·Journal of Marketing, 70 (2007)] its influence in the To test the hypothesis that the influence of marketing on corporate strategy today is different from 2000-2004 period, a random sample of 78 U.S. fims is selected. Of these, 26 firms have a chief marketing officer on their top management team. The test is conducted at a significance level of -.05 Let p be the true proportion of firms with a chief marketing officer currently on their top management team. To cond the hypothesis test, the null and alternative hypotheses are fomulated as: O Ho: p s 0.42; Hs: p>0.42 O Ho: p 2 0.42; Ha: p

Explanation / Answer

5.

As it uses the word "different", then we know it is a two tailed test.

Hence,

Ho: p = 0.42, Ha: p=/= 0.42 [ANSWER, A]

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Here,  
n =    78
p =    0.42

Hence,

u = mean = p =    0.42
  
s = standard deviation = sqrt(p(1-p)/n) =    0.055884496

Hence,

If the null hypothesis is true, the sampling distribution of the sample proportion p^ can be approximated by a ____NORMAL DISTRIBUTION____ with a mean ____0.42___ and a standard deviation of ___0.0558844___. [ANSWERS]

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Formulating the null and alternatuve hypotheses,          
          
Ho:   p   =   0.42
Ha:   p   =/=   0.42
As we see, the hypothesized po =   0.42      

Getting the point estimate of p, p^,          
          
p^ = x / n =    0.333333333      
          
Getting the standard error of p^, sp,          
          
sp = sqrt[po (1 - po)/n] =    0.055884496      
          
Getting the z statistic,          
          
z = (p^ - po)/sp =    -1.550817713   [ANSWER, TEST STATISTIC]  

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As alpha = 0.05, two tailed,

OPTION A: Reject Ho if z < -1.96 or if z > 1.96. [ANSWER]

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As this is a    2   tailed test, then, getting the p value,  
          
p =    0.120945375   [ANSWER, P VALUE]

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LAST PARAGRAPH:

First blank: NOT REJECTED
Second: |z| < 1.96.
Third: NOT REJECTED
Fourth: P > 0.05.
Fifth: CANNOT CONCLUDE

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