Question

# In the month of january researchers have been collecting samples from ten locations for practically all...

In the month of january researchers have been collecting samples from ten locations for practically all days of the month (30 days). The average was reported to be 55 degrees and the standard deviation was 11 degrees. Create:

a) A 95% confidence interval using a normal distribution around the mean.

b) A 90% confidence interval using a normal distribution around the mean.

month of Jamuary researchers have been collecting samples from ten locations for practically all days of the month (30 days). The average was reported to be 55 degrees and the standard deviation was 11 degrees. Create:

a) A 95% confidence interval using a normal distribution

b) A 90% confidence interval using a normal distribution

```What should be the sample average so that you reject the null hypothesis with alpha=0.05?
Assume that all conditions necessary for inference are satisfied. ```
```Whats the Null and Alt of the following?
```

a) Since 2008, chain restaurants in California have been required to display calorie counts of each menu item. Prior to menus displaying calorie counts, the average calorie intake of diners at a restaurant was 1100 calories. After calorie counts started to be displayed on menus, a nutritionist collected data on the number of calories consumed at those restaurants from a random sample of diners. She is trying to determine whether there is convincing evidence of a difference in the average calorie intake for diners at these restaurants.

b) Based on the performance of those who took the GRE exam between July 1, 2004 and June 30, 2007, the average Verbal Reasoning score was calculated to be 462. In 2011 the average verbal score was slightly higher. Do these data provide convincing evidence that the average GRE Verbal Reasoning score has changed since 2004?

Number of samples, n = 10 * 30 = 300

Standard error of the sample mean = SD / = 11 / = 0.6350853

a)

z value for 95% confidence interval = 1.96

Margin of error = z * standard error = 1.96 * 0.6350853 = 1.244767

95% confidence interval is,

(55 - 1.244767, 55 + 1.244767)

= (53.75523, 56.24477)

b)

z value for 90% confidence interval = 1.64

Margin of error = z * standard error = 1.64 * 0.6350853 = 1.04154

95% confidence interval is,

(55 - 1.04154, 55 + 1.04154)

= (53.95846, 56.04154)

For alpha = 0.05, the critical value of Z is 1.96.

Margin of error = z * standard error = 1.96 * 0.6350853 = 1.244767

Confidence interval is,

(55 - 1.04154, 55 + 1.04154)

= (53.95846, 56.04154)

We reject the null hypothesis when the sample average is less than 53.95846 or greater than 56.04154.

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