Using the information below write a pair of hypothesis for e

Using the information below, write a pair of hypothesis for each of the claims and perform the hypothesis test. Be sure to list ?, critical values, standardized test statistics, P-values, and decision. Interpret the results.

With 90% confidence, the average number of red candies is 3.

With 99% confidence, the average number of blue candies is less than 2.5.

With 95% confidence, the variance of the number of green candies is greater than or equal to 2.5.

With 90% confidence, the variance of the number of red candies is 3.

Using the information below, write a pair of hypothesis for each of the claims and perform the hypothesis test. Be sure to list ?, critical values, standardized test statistics, P-values, and decision. Interpret the results. With 90% confidence, the average number of red candies is 3. With 99% confidence, the average number of blue candies is less than 2.5. With 95% confidence, the variance of the number of green candies is greater than or equal to 2.5. With 90% confidence, the variance of the number of red candies is 3.

Solution

With 90% confidence, the average number of red candies is 3.

Null hypothesis: mu=3

Alternative hypothesis: mu not equal to 3

The test statistic is

Z=(xbar-mu)/(s/vn)

=(2.867-3)/(1.61316/sqrt(86))

=-0.76

It is a two-tailed test.

Given a=1-0.9=0.1, the critical values are Z(0.05) =-1.645 or 1.645 (from standard normal table)

The rejection regions are if Z<-1.645 or Z>1.645, we reject the null hypothesis.

Since Z=-0.76 is between -1.645 and 1.645, we do not reject the null hypothesis.

The p-value= 2*P(Z<-0.76) =0.4473 (from standard normal table)

So we can conclude that the average number of red candies is 3.

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With 99% confidence, the average number of blue candies is less than 2.5.

Null hypothesis: mu=2.5

Alternative hypothesis: mu<2.5

The test statisitc is

Z=(xbar-mu)/(s/vn)

=(2.767-2.5)/(1.675036/sqrt(83))

=1.45

Given a=1-0.99=0.01, the critical value is Z(0.01)=-2.33 (from standard normal table)

The rejection region is if Z<-2.33, we reject the null hypothesis.

Since Z=1.45 is lager than -2.33, we do not reject the null hypothesis.

The p-value= P(Z<1.45) =0.9265 (from standard normal table)

So we can not conclude that the average number of blue candies is less than 2.5.

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With 95% confidence, the variance of the number of green candies is greater than or equal to 2.5.

Null hypothesis: o^2=2.5

Alternative hypothesis: o^2>=2.5

The test statisitc is

chisquare= (n-1)*s^2/o^2

=(87-1)*1.955172/2.5

=67.26

The degree of freedom =n-1=87-1=86

Given a=0.05, the critical value of chisquare with 0.95 and df=86 is 108.65 (from chisquare table)

The rejection region is if chsiqure >108.65, we reject the null hypothesis.

Since 67.26 is less than 108.65, we do not reject the null hypothesis.

The p-value= P(X^2 with df=86 >67.26) =0.9327 (from chisquare table)

So we can not conclude that the variance of the number of green candies is greater than or equal to 2.5.

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With 90% confidence, the variance of the number of red candies is 3.

Null hypothesis: o^2=3

Alternative hypothesis: o^2 not equal to 3

The test statisitc is

chisquare= (n-1)*s^2/o^2

=(86-1)*2.60229885/3

=73.73

The degree of freedom =n-1=86-1=85

Given a=0.1, the critical values of chisquare with 0.05, df=85 is 64.75 (from chisquare table)

chisquare with 0.95 and df=85 is 107.52 (from chisquare table)

The rejection regions are if X^2<64.75 or X^2>107.52, we reject the null hypothesis.

Since 73.73 is between 64.75 and 107.52, we do not reject the null hypothesis.

The p-value= P(X^2 with df=85 >73.73) =0.8034 (from chisquare table)

So we can conclude that the variance of the number of red candies is 3.