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Alexus [3.1K]
3 years ago
8

Using the same scenario as above, but now you do not know the population variance. You collect a random sample of n = 21 cereal

boxes and find the sample mean is 12.8 and the sample standard deviation to be 0.5. Test the null hypothesis H0: µ = 13 against the alternative hypothesis HA: µ < 13 at level of significance α = 0.01.
Required:
a. Find the test statistic.
b. Using the rejection region, do you reject or not reject the null hypothesis?
c. Find the p-value.
d. Using the p-value, do you reject or not reject the null hypothesis?
e. In context of your problem, what is your conclusion about the ounces contained in cereal boxes?
Mathematics
1 answer:
padilas [110]3 years ago
3 0

Answer:

-1.833 ; we do not reject the Null.

Pvalue = 0.0409 ; We do not reject the Null

There is no significant evidence to conclude that the number of ounces contained in the cereal boxes is less than 13

Step-by-step explanation:

H0 : μ = 13

HA : μ < 13

Test statistic :

(xbar - μ) ÷ s/sqrt(n)

(12.8 - 13) ÷ 0.5/sqrt(21)

-0.2 / 0.1091089

= - 1.833

The critical value at α = 0.01 ; df = 21 - 1 = 20 = 2.528

-1.833 > - 2. 528 (left tailed test) ; Hence we do not reject the Null.

Test statistic :

Pvalue from Test statistic at DF = 20 ; α = 0.01

Pvalue = 0.0409

Pvalue > α

0.0409 > 0.01 ; Hence, we do not reject H0.

There is no significant evidence to conclude that the number of ounces contained in the cereal boxes is less than 13

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Answer:

Step-by-step explanation:

Solution :  

hello :

note :

Use the point-slope formula.

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Answer:

(-1, -5) satisfies the inequality and will be the solution to the inequality.

From the attached figure, it is clear that  (-1, -5) is below the line.

Therefore, the correct option is:

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Step-by-step explanation:

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Therefore, the correct option is:

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leva [86]

Answer:

We conclude that the actual percentage of households is equal to 30%.

Step-by-step explanation:

We are given that a consumer group suspects that the proportion of households that have three cell phones NOT known to be 30%.

Their marketing people survey 150 households with the result that 43 of the households have three cell phones.

Let p = <u><em>proportion of households that have three cell phones NOT known.</em></u>

So, Null Hypothesis, H_0 : p = 30%      {means that the actual percentage of households is equal to 30%}

Alternate Hypothesis, H_A : p \neq 30%     {means that the actual percentage of households different from 30%}

The test statistics that would be used here <u>One-sample z-test for proportions;</u>

                                T.S. =  \frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of households having three cell phones = \frac{43}{150} = 0.29

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So, <u><em>the test statistics</em></u>  =  \frac{0.29-0.30}{\sqrt{\frac{0.30(1-0.30)}{150} } }  

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The value of z test statistic is -0.27.

<u>Also, P-value of the test statistics is given by;</u>

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Therefore, we conclude that the actual percentage of households is equal to 30%.

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