Question

Data were collected on the amount spent by 64 customers for lunch at a major Houston restaurant. These data are contained in the WEBfile named Houston. Based upon past studies the population standard deviation is known with = $6.Click on the webfile logo to reference the data.Round your answers to two decimal places.a. At 99% confidence, what is the margin of error?b. Develop a 99% confidence interval estimate of the mean amount spent for lunch.____ to ____Amount20.5014.6323.7729.9629.4932.709.2020.8928.8715.7818.1612.1611.2216.4317.669.5918.8919.8823.1120.1120.3420.0830.3621.7921.1819.2234.1327.4936.5518.3732.2712.6325.5327.7133.8121.7919.1626.3520.0126.8513.6317.2213.1720.1222.1122.4720.3635.4711.8517.886.8330.9914.6218.3826.8525.1027.5525.8714.3715.6126.4624.2416.6620.85

Answer 1

Answer:

attached below

Step-by-step explanation:

Answer 2

The 90% CI  by using normal dist is "$19.585 < \mu < 23.455$". A complete solution is below.

According to the table,

The sample mean will be:

• $\bar{x} = \frac{\Sigma (x)}{n}$

           $= \frac{1377.28}{64}$

           $= 21.52$

Standard deviation,

• $\sigma = 6$

Sample size,

• $n = 64$

Significance level,

• $\alpha = 1-CI$

           $= 1-0.99$

           $= 0.01$  

Critical value,

• $z_{\frac{\alpha}{2} } = z_{0.005}$

             $= 2.58$

Now,

The margin of error will be:

→ $E = z_{\frac{\alpha}{2} }\times \frac{\sigma}{\sqrt{n} }$

By substituting the values, we get

       $= 2.58\times \frac{6}{\sqrt{64} }$

       $= 2.58\times \frac{6}{8}$

       $= 1.935$

Limits will be:

→ Lower = $\bar{x}-E$

              = $21.52-1.935$

              = $19.585$

→ Upper = $\bar{x} +E$

              = $21.52+1.935$

              = $23.455$

hence,

The 99% CI is:

= $\bar{x} \pm E$

= $21.52 \pm 1.935$

= $(19.585, 23.455)$

Thus the above answer is correct.

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