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Suppose that an accounting firm does a study to determine the time needed to complete one person's tax forms. It randomly surveys 175 people. The sample average is 23.3 hours. There is a known population standard deviation of 6.2 hours. The population distribution is assumed to be normal. NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) Construct a 90% confidence interval for the population average time to complete the tax forms

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JeanaShupp

Answer: (22.53, 24.07)

Step-by-step explanation:

As per given we have,

n= 175 , [tex]\overline{x}=23.3[/tex]  , [tex]\sigma=6.2[/tex]

Critical value for 90% confidence interval : [tex]z_{\alpha/2}=1.645[/tex]

The population distribution is assumed to be normal.

Since population standard deviation is known , so we use z-test.

Confidence interval :

[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]

[tex]23.3\pm (1.645)\dfrac{6.2}{\sqrt{175}}\\\\=23.3\pm0.77\\\\=(23.3-0.77,\ 23.3+0.77)\\\\=(22.53,\ 24.07)[/tex]

Hence,  a 90% confidence interval for the population average time to complete the tax forms = (22.53, 24.07)

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