Consider the following hypotheses: H 0 : ? = 4 , 900 H A : ? ? 4 , 900 The population is...

Question:

Consider the following hypotheses:

{eq}H0: \mu = 4,900 {/eq}

{eq}HA: \mu \neq 4,900 {/eq}

The population is normally distributed with a population standard deviation of 600. Compute the value of the test statistic and the resulting p-value for each of the following sample results. For each sample, determine if you can "reject/do not reject" the null hypothesis at the 10% significance level.

Test Statistic p-value Respond with: Reject H0 or Do not reject H0
a x (bar) =4,960; n = 125
b x (bar) =4,960; n = 305
c X (bar) = 4,740; n = 36
d X(bar)= 4,770; n = 36

Two-tailed Hypothesis Testing:

Hypothesis testing, also called significance testing, is a process carried out by statistical analyst to determine whether null hypothesis should be rejected or not to be rejected. The null hypothesis is the existing evidence while alternative hypothesis is the researcher's claim or opinion.

Answer and Explanation: 1

Test Statistic p-value Respond with: Reject H0 or Do not reject H0
a x (bar) =4,960; n = 125 1.12 0.1314Do not reject the null hypothesis
b x (bar) =4,960; n = 305 1.25 0.0401 Reject the null hypothesis
c X (bar) = 4,740; n = 36 -1.6 0.0548 Do not reject the null hypothesis
d X(bar)= 4,770; n = 36 -1.3 0.0968 Do not reject the null hypothesis

The calculations are done below:


a).

Since the distribution is normally distributed, we'll use standard normal distribution:

{eq}\begin{align*} \displaystyle z&=\frac{\bar X-\mu} {\displaystyle \frac{\sigma}{\sqrt{n}}}\\\displaystyle &=\frac{4960-4900}{\displaystyle \frac{600}{\sqrt{125}}}\\&=1.12 \end{align*} {/eq}

Since the the hypothesis is a two tailed test, the p-value will be on both sides of the curve:

{eq}p\text{-value}=P(z<-1.12)\,\text{or}\, P(z>1.12)\\P(z<-1.12)=0.1314 {/eq}

Since p-value is greater than level of significance, we reject the null hypothesis.


b).

{eq}\begin{align*} \displaystyle z&=\frac{4960-4900}{\displaystyle \frac{600}{\sqrt{305}}}\\&=1.75 \end{align*} {/eq}

{eq}P\text{-value}=P(z<-1.75)=P(z>1.75)\\P(z<-1.75)=0.0401\\\therefore p\text{-value}=0.0401 {/eq}

Reject the null hypothesis.


c).

{eq}\begin{align*} \displaystyle z&=\frac{4740-4900}{\displaystyle \frac{600}{\sqrt{36}}}\\&=-1.6 \end{align*} {/eq}

{eq}P\text{-value}=P(z<-1.6)=P(z>1.6)\\P(z<-1.6)=0.0548\\\therefore p\text{-value}=0.0548 {/eq}

Do not reject the null hypothesis. This is because p-value is greater than level of significance.


d).

{eq}\begin{align*} \displaystyle z&=\frac{4770-4900}{\displaystyle \frac{600}{\sqrt{36}}}\\&=-1.3 \end{align*} {/eq}

{eq}P\text{-value}=P(z<-1.3)=P(z>1.3)\\P(z<-1.3)=0.0968\\\therefore p\text{-value}=0.0968 {/eq}


Learn more about this topic:

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Two-Tailed Test: Formula & Examples

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Chapter 20 / Lesson 16
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Learn what a two-tailed test is in statistics. Understand how to conduct a two-tailed hypothesis test using certain formulas from a two-tailed test example.


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