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melamori03 [73]
3 years ago
8

In one area, monthly incomes of technology-related workers have a standard deviation of $650. It is believed that the standard d

eviation of the monthly incomes of non-technology workers is different. 71 non-technology workers are randomly selected and it is determined that these selected workers have a standard deviation of $950. Test the claim that the non-technology workers have a different standard deviation (so different from $650). Use a 0.10 significance level.
Mathematics
1 answer:
Virty [35]3 years ago
6 0

Answer:

There is sufficient statistical evidence to prove that the standard deviation of the technology-related workers and the standard deviation of the non-technology workers are equal.

Step-by-step explanation:

Here we have our null hypothesis as H₀: σ² = s²

Our alternative hypothesis is then Hₐ: σ² ≠ s²

We therefore have a two tailed test

To test the hypothesis of difference in standard deviation which is the Chi squared test given as follows

\chi ^{2} = \dfrac{\left (n-1  \right )s^{2}}{\sigma ^{2}}

Where:

n = Size of sample

s² = Variance of sample = 950²

σ² = Variance of population = 650²

Degrees of freedom = n - 1 = 71 - 1 = 70

α = Significance level = 0.1

Therefore, we use 1 - 0.1 = 0.9

From the Chi-square table, we have the critical value as

1 - α/2 = 51.739,  

α/2 = 90.531

Plugging the values in the above Chi squared test equation, we have;

\chi ^{2} = \dfrac{\left (23-1  \right )950^{2}}{650 ^{2}} = 49.994

Therefore, since the test value within the critical region, we do not reject the null hypothesis, hence there is sufficient statistical evidence to prove that the standard deviation of the technology-related workers and the standard deviation of the non-technology workers are equal.

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Step-by-step explanation:

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Calculating 25% of $9.50 = \frac{25}{100}  \times 9.50 = 2.375

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Hence, the total amount of money he has to pay for the dish is $8.3861.

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