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nlexa [21]
3 years ago
15

A simple random sample of 28 Lego sets is obtained and the number of pieces in each set was counted.The sample has a standard de

viation of 12.65. Use a 0.05 significance level to test the claim that the number of pieces in a set has a standard deviation different from 11.53.
Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
6 0

Answer:

Step-by-step explanation:

Given that:

A simple random sample n = 28

sample standard deviation S = 12.65

standard deviation \sigma = 11.53

Level of significance ∝ = 0.05

The objective is to test the claim that the number of pieces in a set has a standard deviation different from 11.53.

The null hypothesis and the alternative hypothesis can be computed as follows:

Null hypothesis:

H_0: \sigma^2 = \sigma_0^2

Alternative hypothesis:

H_1: \sigma^2 \neq \sigma_0^2

The test statistics can be determined by using the following formula in order to test if the claim is statistically significant or not.

X_0^2 = \dfrac{(n-1)S^2}{\sigma_0^2}

X_0^2 = \dfrac{(28-1)(12.65)^2}{(11.53)^2}

X_0^2 = \dfrac{(27)(160.0225)}{132.9409}

X_0^2 = \dfrac{4320.6075}{132.9409}

X_0^2 = 32.5002125

X^2_{1- \alpha/2 , df} = X^2_{1- 0.05/2 , n-1}

X^2_{1- \alpha/2 , df} = X^2_{1- 0.025 , 28-1}

From the chi-square probabilities table at 0.975 and degree of freedom 27;

X^2_{0.975 , 27} = 14.573

X^2_{\alpha/2 , df} = X^2_{ 0.05/2 , n-1}

X^2_{\alpha/2 , df} = X^2_{0.025 , 28-1}

From the chi-square probabilities table at 0.975 and degree of freedom 27;

X^2_{0.025 , 27}= 43.195

Decision Rule: To reject the null hypothesis if X^2_0  \ >  \ X^2_{\alpha/2 , df}  \ \  \ or \ \ \   X^2_0 \  < \  X^2_{1- \alpha/2 , df} ; otherwise , do not reject the null hypothesis:

The rejection region is X^2_0  \ >  43.195 \ \  \ or \ \ \   X^2_0 \  < \  14.573

Conclusion:

We fail to reject the null hypothesis since  test statistic value 32.5002125  lies  between 14.573 and 43.195.

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