Question

..NEED ANSWER ASAP / need correct answers for following

A business journal investigation of the performance and timing of corporate acquisitions discovered that in a random sample of 2.984 firms, 843 announced one or more acquisitions during the year 2000. Does the sample provide sufficient evidence to indicate that the true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 31%? Use a = 0.05 to make your decision. What are the hypotheses for this test? O A. He: p=0.31 Hyp*0.31 O C. Ho: p=0.31 Ha: p<0.31 O E. Ho p>0.31 HpS0.31 O B. Ho p70.31 He: p=0.31 OD. Ho: p<0.31 Hy p=0.31 OF. Ho p=0.31 He p>0.31 What is the rejection region? Select the correct choice below and fill in the answer box(es) to complete your choice. (Round to three decimal places as needed.) O A. z> OB. z< orz> OC. z< Calculate the value of the 2-statistic for this test.
reject, do not reject , is , is not, sufficient or insufficient

Answer 1

First problem:

(C)

H₀: p = 0.31
H₁: p < 0.31

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For 0.05 significance level, z = Normsinv(0.05) = -1.645

So, the rejection region is:

(C)

z < -1.645

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Calculation for the test statistic (z)

p̂ = 843 / 2984 = 0.2825

z = (p̂ - 0.31) / SQRT(0.31*(1 - 0.31)/2984) = (0.2825 - 0.31) / SQRT(0.31*(1 - 0.31)/2984) = -3.248

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Note that z = -3.248 is less than -1.645.

So,

Reject the null hypothesis because the test statistic is in the rejection region. Therefore there is sufficient evidence that the true % is less than 31%.

Second problem:

x̄ = 10.7
s = 3.2
n = 50

The sample size is more than 30 which enables the test statistic to follow normal distribution. Otherwise, a t-distribution can always be used.

Test statistic, t = (x̄ - 10) / s√n = (10.7 - 10) / (3.2/SQRT(50)) = 1.547

Corresponding p-value = 2*T.DIST.RT(1.547,50-1) = 0.1283 (if t-distribution is used)
Corresponding p-value = 2*(1 - NORMSDIST(1.547)) = 0.1219 (if normal distribution is used)

A)

There is sufficient evidence to reject H0 for alpha > 0.13 because the p-value of 0.1283 or 0.1219 is less than alpha.