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Of a random sample of 200 auditors, 105 indicated some measure of agreement with this statement: "Cash flow is an important indicator of profitability." [Assume the sample is large](a)Test at the 10% significance level against a two-sided alternative the null hypothesis that one-half of the members of this population would agree with this statement. (b)Find and interpret the p-value of this test.

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Answer:

a)At the 10% significance level, we fail to reject the null hypothesis.

b) P-value is ≈ 0.48

Step-by-step explanation:

Let p be the proportion of the auditors agree with the statement  "Cash flow is an important indicator of profitability."

Then,

[tex]H_{0}[/tex]: p=0.5

[tex]H_{a}[/tex]: p≠0.5

Test statistic can be calculated as:

z=[tex]\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }[/tex] where

  • p(s) is the sample proportion of the auditors agree with the statement. ([tex]\frac{105}{200}[/tex]=0.525)
  • p is the proportion assumed under null hypothesis. (0.5)
  • N is the sample size (200)

Using these numbers we get:

z=[tex]\frac{0.525-0.5}{\sqrt{\frac{0.5*0.5}{200} } }[/tex]≈0,707

and p(z) is ≈ 0.48. Since 0.48>0.10, the result is not significant at 90% level. Therefore we fail to reject the null hypothesis.