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mina [271]
3 years ago
12

Bastien, Inc. has been manufacturing small automobiles that have averaged 50 miles per gallon of gasoline in highway driving. Th

e company has developed a more efficient engine for its small cars and now advertises that its new small cars average more than 50 miles per gallon in highway driving. An independent testing service road-tested 36 of the automobiles. The sample showed an average of 51.5 miles per gallon. The population standard deviation is 6 miles per gallon. You conduct a test with a 0.05 level of significance to determine whether or not the manufacturer's advertising campaign is legitimate. What is the p-value associated with the sample results (2 decimals)?
Mathematics
1 answer:
alexira [117]3 years ago
3 0

Answer:

We conclude that the the manufacturer's advertising campaign is not correct.

P-value of test = 0.067.

Step-by-step explanation:

We are given that an independent testing service road-tested 36 of the automobiles. The sample showed an average of 51.5 miles per gallon. The population standard deviation is 6 miles per gallon.

We have to conduct a test to determine whether or not the manufacturer's advertising campaign is legitimate.

<u><em>Let </em></u>\mu<u><em> = population average gasoline of new small cars in highway driving.</em></u>

SO, <u>Null Hypothesis</u>, H_0 : \mu \leq 50 miles per gallon   {means that the manufacturer's advertising campaign is not correct}

<u>Alternate Hypothesis</u>, H_A : \mu > 50 miles per gallon   {means that the manufacturer's advertising campaign is correct}

The test statistics that will be used here is <u>One-sample z test</u> <u>statistics</u> as we know about the population standard deviation;

                         T.S.  = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where,  \bar X = sample average gasoline = 51.5 miles per gallon

              \sigma = population standard deviation = 6 miles per gallon

              n = sample of automobiles = 36

So, <em><u>test statistics</u></em>  =   \frac{51.5-50}{\frac{6}{\sqrt{36} } }

                               =  1.50

<em>Now at 0.05 significance level, the z table gives critical value of 1.6449 for right-tailed test. Since our test statistics is less than the critical values of z as 1.50 < 1.6449, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.</em>

<u>Also, P-value is given by the following formula;</u>

         P-value = P(Z > 1.50) = 1 - P(Z \leq 1.50)

                                            = 1 - 0.93319 = <u>0.067 or 6.7%</u>

Therefore, we conclude that the the manufacturer's advertising campaign is not correct as its new small cars average less than or equal to 50 miles per gallon in highway driving.

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