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Business Statistics

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4. A survey camied out by John claims that the average cost of living for households in City 1 is more than that in City 2 :

$$ \begin{array}{|c|c|} \hline \text { City } 1 & \text { City } 2 \\ \hline \overline{x_{1}}=164 & \overline{x_{2}}=159 \\ \hline \sigma_{1}=12.50 & \sigma_{2}=9.25 \\ \hline n_{1}=35 & n_{2}=30 \\ \hline \end{array} $$

(a) Specify the competing hypotheses to test John's claim.

(b) Calculate the value of the test statistic.

(c) At the \(5 \%\) significance level, is John's claim supported by the data? Explain

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Answer #1

The provided sample means are shown below:

$$ \begin{array}{l} \bar{X}_{1}=164 \\ \bar{X}_{2}=159 \end{array} $$

Also, the provided population standard deviations are:

$$ \begin{array}{l} \sigma_{1}=12.50 \\ \sigma_{2}=9.25 \end{array} $$

and the sample sizes are \(n_{1}=35\) and \(n_{2}=30\).

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

$$ \begin{array}{l} H o: \mu_{1}=\mu_{2} \\ H a: \mu_{1}>\mu_{2} \end{array} $$

This corresponds to a right-tailed test, for which a z-test for two population means, with known population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is \(\alpha=0.05\), and the critical value for a right-tailed test is \(z_{c}=1.64\)

The rejection region for this right-tailed test is \(R=\{z: z>1.64\}\)

(3) Test Statistics

The z-statistic is computed as follows:

$$ z=\frac{X_{1}-X_{2}}{\sqrt{\sigma_{1}^{2} / n_{1}+\sigma_{2}^{2} / n_{2}}}=\frac{164-159}{\sqrt{12.50^{2} / 35+9.25^{2} / 30}}=1.849 $$

(4) Decision about the null hypothesis

Since it is observed that \(z=1.849>z_{c}=1.64\), it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is \(p=0.0323,\) and since \(p=0.0323<0.05,\)it is concluded that the null hypothesis is rejected.

answered by: BradelNumero
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Answer #2

Given that City-l City - 2 Sample size n₂=30 sample size ni-35 mean ał = 164 mean T2 = 159 2=9.25 S.D Г r = 12.50 The sampleb) Test statistic al - R2 Z + 62 nz 164 -159 (12.5) 9.259 35 30 S 2.7049 1o8485 c) = 1.645 crite cal value of Z at x = 5% [us

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