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Please answer showing all calculations, & state the technique used if applicable

ID: 3291472 • Letter: P

Question

Please answer showing all calculations, & state the technique used if applicable. Thankyou.

Tests carried out on behalf of the Arborists Association on a random sample of twelve telescopic extension ladders manufactured by Safe GardenWorks yielded the following information.

X = 2,700 .0kgs x^2= 609,975.0 kgs^2

Here X is the maximum load that the ladder can support before it shows some sign of deformation.

(a) Find the 99% confidence limits of (i) the true mean and (ii) the true standard deviation of the probability distribution of X . Assume that X is a normally distributed random variable.

(b) Test the hypothesis that the mean value of the probability distribution of X is not significantly different from 220kgs,. Let = 0.05.

(c) Test the hypothesis that the true standard deviation of the probability distribution of X is not significantly different from 10kgs. Let = 0.05.

Explanation / Answer

Given that,

n = 12

X = 2,700 .0kgs x^2= 609,975.0 kgs^2

Here X is the maximum load that the ladder can support before it shows some sign of deformation.

First we have to find mean and standard deviation using these values.

Mean (Xbar) = X/n = 2700/12 = 225

variance = x^2 / n - mean2 = 609975.0/12 - 225^2 = 206.25

standard deviation =sqrt(variance) = sqrt(4827.27) = 14.36

(a) Find the 99% confidence limits of (i) the true mean and (ii) the true standard deviation of the probability distribution of X . Assume that X is a normally distributed random variable.

We can find 99% confidence interval for true mean by using TI-83 calculator.

steps :

STAT --> TESTS --> 8:TInterval --> ENTER --> Highlight on STats --> ENTER --> Input all the values --> C-level : 99.0 --> Calculate --> ENTER

99% confidence interval for true mean is (212.13,237.87).

Now we have to find 99% confidence interval for true standard deviation.

The 99% confidence interval for sigma is,

sqrt[(n-1)*s^2 / X2u ] < sigma < sqrt[(n-1)*s^2 / X2l ]

where X2u and X2l arecritical values for chi square distribution.

We can find critical values by using EXCEL.

syntax :

For X2u :

=CHIINV(probability,deg_freedom)

probability = (1 - C)/2

where C is confidence level = 99.0

deg_freedom = n-1 = 12-1 = 11

FOr X2l :

=CHIINV(probability,deg_freedom)

probability = (1 + C)/2

deg_freedom = 11

X2u = 26.76

X2l = 2.60

sqrt[(n-1)*s^2 / X2u ] < sigma < sqrt[(n-1)*s^2 / X2l ]

sqrt [(11*14.362)/26.76] < sigma < sqrt[(11*14.362) / 2.60]

9.21< sigma < 29.54

The 99% confidence interval for standard deviation is (9.21, 29.54)

(b) Test the hypothesis that the mean value of the probability distribution of X is not significantly different from 220kgs,. Let = 0.05.

Now we have to test the hypothesis that,

H0 : mu = 220 Vs H1 : mu not= 220

where mu is population mean.

Here we use one sample t-test.

We can do one sample t-test in TI-83 calculator.

steps :

STAT --> TESTS --> 2: TTest --> ENTER --> Highlight on Stats --> ENTER --> Input all the values --> Alternative : not equal --> Calculate --> ENTER

test statistic = 1.2062

P-value = 0.2531

P-value > alpha

Accept H0 at 5% level of significance.

Conclusion : There is not sufficient evidence to say that the mean value of the probability distribution of X is not significantly different from 220kgs.

c) Test the hypothesis that the true standard deviation of the probability distribution of X is not significantly different from 10kgs. Let = 0.05.

Here we have to test the hypothesis that,

H0 : sigma = 10 vs H1 : sigma not= 10

Here test statistic follows Chi-square distribution with n-1 dgerees of freedom.

The test statistic is,

X2 = (n-1) * s2 / sigma2

= (12-1)* / 14.362 / 102

= 22.69

Now we have to find p-value for taking decision.

P-value we can find in EXCEL.

syntax :

=CHIDIST(x, deg_freedom)

where x is test statistic

deg_freedom = n-1

P-value = 0.020

P-value < alpha

Reject H0 at 5% level of significance.

Conclusion : There is sufficient evidence to say that the true standard deviation of the probability distribution of X is not significantly different from 10kgs.