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jek_recluse [69]
3 years ago
6

The following 5 questions are based on this information: An economist claims that average weekly food expenditure of households

in City 1 is more than that of households in City 2. She surveys 35 households in City 1 and obtains an average weekly food expenditure of $164. A sample of 30 households in City 2 yields an average weekly expenditure of $159. Historical data reveals that the population standard deviation for City 1 and City 2 are $12.50 and $9.25, respectively.
Required:
At the 5% significance level, is the economist claim supported by the data?
Mathematics
1 answer:
Gnom [1K]3 years ago
4 0

Answer:

>> Null hypothesis; H0: μ1 - μ2 ≥ 0

Null hypothesis; Ha: μ1 - μ2 < 0

>> z = 1.85

>> p-value = 0.03

>> there is evidence sufficient to reject the claim that average weekly food expenditure of households in City 1 is more than that of households in City 2.

The economist claim is not supported by data.

Step-by-step explanation:

We are given;

Sample mean of city 1; x1¯ = 164

Sample mean of city 2; x2¯ = 159

Sample size of city 1; n1 = 35

Sample size of city 2; n2 = 30

Standard deviation of city 1; σ1 = 12.5

Standard deviation of city 2; σ2 = 9.25

Let's define the hypotheses;

Null hypothesis; H0: μ1 - μ2 ≥ 0

Null hypothesis; Ha: μ1 - μ2 < 0

Test statistic which is the z-score Formula between 2 mean is;

z = ((x1¯ - x2¯) - 0)/√((σ1)²/n1) + ((σ2)²/n2))

z = (164 - 159 - 0)/√((12.5²/35) + (9.25²/30))

z = 5/2.705

z = 1.85

From online p-value from z-score calculator attached, using z = 1.85, one tailed, significance value of 0.05, we have;

p-value = 0.03

The p-value is less than the significance value, and so we will reject the null hypothesis and conclude that there is evidence sufficient to reject the claim that average weekly food expenditure of households in City 1 is more than that of households in City 2.

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A researcher would like to determine whether a new tax on cigarettes has had any effect on people’s behavior. During the year be
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Answer:

a) p_v =2*P(z  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the population mean is NOT significant different from 410 at 5% of significance.  

b) p_v =2*P(z  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the population mean is significant different from 410 at 5% of significance.

c) The expplanation why we have different outcomes is because for part a we use a higher standard error compared to part b. So we have enough evidence on part b to reject the null hypothesis that we no have significant difference from 410.

Step-by-step explanation:

1) Part a

Previous concepts  and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=386 represent the sample mean  

\sigma=60 represent the population standard deviation

n=9 represent the sample selected

\alpha=0.05 significance level  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if we have significant difference on the mean, the system of hypothesis would be:  

Null hypothesis:\mu = 410  

Alternative hypothesis:\mu \neq 410  

If we analyze the size for the sample is < 30 but we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

z=\frac{386-410}{\frac{60}{\sqrt{9}}}=-1.2    

P-value

Since is a two side test the p value would be:  

p_v =2*P(z  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the population mean is NOT significant different from 410 at 5% of significance.  

2) Part b

State the null and alternative hypotheses.  

The system of hypothesis not changes:

Null hypothesis:\mu = 410  

Alternative hypothesis:\mu \neq 410  

Same statistic:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

Calculate the statistic

But now the population deviation changes \sigma=30. We can replace in formula (1) the info given like this:  

z=\frac{386-410}{\frac{30}{\sqrt{9}}}=-2.4    

P-value

Since is a two side test the p value would be:  

p_v =2*P(z  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the population mean is significant different from 410 at 5% of significance.

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