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Bezzdna [24]
3 years ago
10

Consider the null hypothesis, H0: µ = 4,000, at the 1% level of significance. If the z-test statistic is calculated to be 6.00,

which of the following would be the correct decision regarding the null hypothesis?a. Do not reject H0b. reject H0c. Reject H1d. None apply
Mathematics
1 answer:
kow [346]3 years ago
7 0

Answer:

Option B) Reject null hypothesis      

Step-by-step explanation:

We are given the following in the question:

We are given the null hypothesis:

H_0 = 4,000

\alpha = 0.01

z_{stat} = 6.00

Two tailed z-test

Now, z_{critical} \text{ at 0.01 level of significance } = \pm 2.58

Since,  

The calculated z-statistic does not lie in the acceptance region, we fail to accept and reject the null hypothesis.

Left-tailed z-test

Now, z_{critical} \text{ at 0.01 level of significance } = -2.58

Since,  

z_{stat} > z_{critical}

We fail to accept and reject the null hypothesis.

Right-tailed z-test

Now, z_{critical} \text{ at 0.01 level of significance } = 2.58

Since,  

z_{stat} > z_{critical}

We fail to accept and reject the null hypothesis.

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\bf ~\hspace{5em} \textit{ratio relations of two similar shapes} \\\\ \begin{array}{ccccllll} &\stackrel{\stackrel{ratio}{of~the}}{Sides}&\stackrel{\stackrel{ratio}{of~the}}{Areas}&\stackrel{\stackrel{ratio}{of~the}}{Volumes}\\ \cline{2-4}&\\ \cfrac{\stackrel{similar}{shape}}{\stackrel{similar}{shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}~\hspace{6em} \cfrac{s}{s}=\cfrac{\sqrt{Area}}{\sqrt{Area}}=\cfrac{\sqrt[3]{Volume}}{\sqrt[3]{Volume}} \\\\[-0.35em] \rule{34em}{0.25pt}

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