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.

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,

: p=0.5

: p≠0.5

Test statistic can be calculated as:

z= where

Using these numbers we get:

z=≈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.

answer: he is 272195

step-by-step explanation: all you have to do is multiple 3535 and 77 and it equals 272195

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