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According to the local union president, the mean gross income of plumbers in the Salt Lake City area follows a normal distribution with a mean of $48,000 and a population standard deviation of $2,000. A recent investigative reporter for KYAK TV found, for a sample of 49 plumbers, the mean gross income was $47,600. At the 0.05 significance level, is it reasonable to conclude that the mean income is not equal to $47,600? Determine the p value. State the Null and Alternate hypothesis: State the test statistic: State the Decision Rule: Show the calculation: What is the interpretation of the sample data? Show the P value

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ajeigbeibraheem

Answer:

Step-by-step explanation:

Given that:

population mean [tex]\mu[/tex] = 47600

population standard deviation [tex]\sigma[/tex] = 2000

sample size n = 49

Sample mean [tex]\over\ x[/tex] = 48000

Level of significance = 0.05

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

[tex]H_0 : \mu = 47600 \\ \\ H_1 : \mu \neq 47600[/tex]

Using the table of standard normal distribution, the value of z that corresponds to the two-tailed probability 0.05 is 1.96. Thus, we will reject the null hypothesis if the value of the test statistics is less than -1.96 or more than 1.96.

The test statistics can be calculated by using the formula:

[tex]z= \dfrac{\overline X - \mu }{\dfrac{\sigma}{ \sqrt{n}}}[/tex]

[tex]z= \dfrac{ 48000-47600 }{\dfrac{2000}{ \sqrt{49}}}[/tex]

[tex]z= \dfrac{400 }{\dfrac{2000}{ 7}}[/tex]

[tex]z= 1.4[/tex]

Conclusion:

Since 1.4 is lesser  than 1.96 , we fail to reject the null hypothesis and  that there is insufficient information to conclude that the   mean gross income is not equal to $47600

The P-value is being calculate as follows:

P -value = 2P(Z>1.4)

P -value =  2 (1 - P(Z< 1.4)

P-value = 2 ( 1 - 0.91924)

P -value = 2 (0.08076 )

P -value = 0.16152

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