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A business journal investigation of the performance and timing of corporate acquisitions discovered that in a random sample of 2,684 ​firms, 715 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 29​%? Use alpha equals0.05 to make your decision.

Calculate the value of the​ z-statistic for this test.

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Answer:

z = 1.960

Step-by-step explanation:

The sample proportion is:

p = 715 / 2684 = 0.2664

The standard error is:

σ = √(pq/n)

σ = √(0.266 × 0.734 / 2684)

σ = 0.0085

For α = 0.05, the confidence level is 95%.  The z-statistic at 95% confidence is 1.960.

The margin of error is 1.960 × 0.0085 = 0.0167.

The confidence interval is 0.2664 ± 0.0167 = (0.2497, 0.2831).

The upper limit is 28.3%, so the journal can conclude with 95% confidence that the true percentage is less than 29%.

Yes, the considered sample provides sufficient evidence to indicate that the true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 29​%.

  • The z-test statistic came out to be -2.69

What is the z test statistic for one sample proportion?

Suppose that we have:

  • n = sample size
  • [tex]\hat{p}[/tex] = sample proportion
  • p = population proportion (hypothesised)

Then, the z test statistic for one sample proportion is:

[tex]Z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]

For this case, we're provided that:

  • Size of sample = n = 2684
  • Sample proportion of those who announced one or more acquisitions during year 2000 is: [tex]\hat{p}[/tex] = 715/n = 715/2684
  • Level of significance = 0.05

We want to determine if true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 29​% = 0.29 (converted percent to decimal).

Hypotheses:

  1. Null hypothesis: [tex]H_0: p \geq p_0 = 0.29\\[/tex] (it nullifies what we want to test for, thus assumes that true mean of the population proportion is not less than 0.29).
  2. Alternate hypothesis: [tex]H_A: p < p_0 = 0.29[/tex] Assumes that the true population mean is less than 0.29

Thus, the test is left tailed test.

where [tex]p_0[/tex] = 29% = 0.29 is the hypothesized mean value of population proportion.

The test statistic is:

[tex]Z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\\\ Z = \dfrac{715/2684 - 0.29}{\sqrt{\dfrac{0.29(1-0.29)}{2684}}} \approx -2.69[/tex]

The critical value of Z at level of significance 0.05 is -1.6449

Since the test statistic = -2.69 < critical value = -1.6449, so the test statistic lies in the rejection region (the rejection region for the left tailed test is all the values below critical value).

Thus, we reject the null hypothesis and accept the alternative hypothesis that the true population mean is less than 0.29.

Thus, the considered sample provides sufficient evidence to indicate that the true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 29​%.

The z-test statistic came out to be -2.69

Learn more about one-sample z-test for population proportion mean here:

https://brainly.com/question/16173147

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