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otez555 [7]
3 years ago
10

Consider the following hypothesis test. H0: μ ≥ 55 Ha: μ < 55 A sample of 36 is used. Identify the p-value and state your con

clusion for each of the following sample results. Use α = 0.01.
(a) x = 54 and s = 5.3 Find the value of the test statistic. (Round your answer to three decimal places.)
Find the p-value. (Round your answer to four decimal places.)
State your conclusion.
(b) x = 53 and s = 4.6
Find the value of the test statistic. (Round your answer to three decimal places.)
Find the p-value. (Round your answer to four decimal places.)
p-value =
(c) x = 56 and s = 5.0
Find the value of the test statistic.
Find the p-value. (Round your answer to four decimal places.)
p-value =
State your conclusion.
Mathematics
1 answer:
Ket [755]3 years ago
6 0

Answer:

Step-by-step explanation:

Given that:

H_o: \mu \ge 55 \ \\ \\ H_1 : \mu < 55

(a) For x = 54 and s = 5.3

The test statistics can be computed as:

Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }

Z = \dfrac{54-55}{\dfrac{5.3}{\sqrt{36}} }

Z = \dfrac{-1}{\dfrac{5.3}{6} }

Z = -1.132

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(-1.132,35,1) = 0.1326

Decision: p-value is greater than significance level; do not reject \mathbf{H_o}

Conclusion: \  There  \ is \  insufficient \  evidence  \  to \  conclude \  that \ \mu < 55

b

For x = 53 and s = 4.6

The test statistics can be computed as:

Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }

Z = \dfrac{53-55}{\dfrac{4.6}{\sqrt{36}} }

Z = \dfrac{-2}{\dfrac{4.6}{6} }

Z = -2.6087

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(-2.6087,35,1) =0.0066

Decision: p-value is < significance level; we reject the null hypothesis.

Conclusion: \  There  \ is \  sufficient \  evidence  \  to \  conclude \  that \mu < 55

c)

For x = 56 and s = 5.0

The test statistics can be computed as:

Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }

Z = \dfrac{56-55}{\dfrac{5.0}{\sqrt{36}} }

Z = \dfrac{56-55}{\dfrac{5.0}{\sqrt{36}} }

Z = 1.2

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(1.2,35,1) = 0.88009

Decision: p-value is greater than significance level; do not reject \mathbf{H_o}

Conclusion: \  There  \ is \  insufficient \  evidence  \  to \  conclude \  that \ \mu < 55

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