Question

Q Search this course According to a recent study annual per capita consumption of milk in the United States is 22.6 gallons. Being from the Midwest, you believe milk consumption is higher there and wish to test your hypothesis. A sample of 14 Online file below. Use the data to set up your spreadsheet and test your hypothesis. from the Midwestern town of Webster City was selected and then each person's milk consumption was entered into the Microsoft Excel a. Develop a hypothesis test that can be used to determine whether the mean annual consumption in Webster City is higher than the national mean. b. What is a point estimate of the difference between mean annual consumption in Webster City and the national mean? (2 decdmals) c. At-0.05, test for a significant difference by completing the following. Caloulate the value of the test statistic (2 decimals). The p-v Reject the nul

Answer 1

Solution:

     We are given that: sample of 14 individuals from the Midwestern town of Webster city and each persons milk consumption was recorded.
     We have to test if mean annual milk consumption is higher than the National mean = 22.6 gallons.

Thus we use following steps:

Part a) State H0 and Ha, to test mean annual milk consumption is higher than the National mean = 22.6 gallons.

$H_{0}: \mu = 22.6$

$H_{a}: \mu > 22.6$

Part b) The Point estimate of the difference between mean annual consumption in Webster city and the national mean.

The point estimate of difference = $\bar{x}- \mu$

where

$\bar{x} = \frac{\sum x}{n}$

Thus we need to make following table:

Gallons of Milk
28.7
23.84
25.25
21.1
17.52
19.61
19.83
26.18
34.97
30
28.59
20.57
26.94
27.24
$\sum x =350.34$

Thus sample mean is:
$\bar{x} = \frac{\sum x}{n}$

$\bar{x} = \frac{350.34 }{14}$

$\bar{x} = 25.02$

Thus

The point estimate of difference = $\bar{x}- \mu$

The point estimate of difference = $25.02-22.6$

The point estimate of difference = $2.42$

Part c) At $\alpha = 0.05$ , test for a significant difference by completing following steps.
Calculate test statistic.

Since population standard deviation is unknown and sample size is small, we use t test statistic.

$t=\frac{\bar{x}-\mu }{s /\sqrt{n}}$

where

$s=\sqrt{\frac{\sum x^{2}-(\sum x)^{2}/n}{n-1}}$

Thus we need to make following table:

x : Gallons of Milk	x^2
28.7	823.6900
23.84	568.3456
25.25	637.5625
21.1	445.2100
17.52	306.9504
19.61	384.5521
19.83	393.2289
26.18	685.3924
34.97	1222.9009
30	900.0000
28.59	817.3881
20.57	423.1249
26.94	725.7636
27.24	742.0176
$\dpi{100} \sum x= 350.34$	$\dpi{100} \sum x^{2}=9076.1270$

Thus sample standard deviation is:

$s=\sqrt{\frac{\sum x^{2}-(\sum x)^{2}/n}{n-1}}$

$s=\sqrt{\frac{9076.1270 -(350.34 )^{2}/14}{14-1}}$

$s=\sqrt{\frac{9076.1270 -8767.0083 }{13}}$

$s=\sqrt{\frac{309.1187 }{13}}$

$s=\sqrt{23.77836 }$

$s=4.8763$

Thus t test statistic is:

$t=\frac{\bar{x}-\mu }{s /\sqrt{n}}$

$t=\frac{2.42 }{4.8763 /\sqrt{14}}$

$t=\frac{2.42 }{4.8763 /3.741657 }$

$t=\frac{2.42 }{1.303248 }$

$t=1.86$

The P-value is:

To find P-value , we use Excel command:

Since Ha is > type, so this is right tailed test, thus we use following excel command:
=T.DIST.RT(x , df )

where x = t = 1.86 and df = n - 1 = 14 - 1 = 13

Thus

=T.DIST.RT(1.86,13)

=0.0428

Thus P-value = 0.0428

Reject the Null Hypothesis:

Yes

Since P-value = 0.0428 < 0.05 significance level.

What is your conclusion?

Since we have rejected null hypothesis H0, there is sufficient evidence to conclude that the mean annual milk consumption is higher there.