Question

1. A climate study compared temperature differences in 1948 vs. 2018. Scientists sampled 197 locations from the National Oceanic and Atmospheric Administration’s (NOAA) historical data, where the data was available for both years of interest. The scientists wanted to know: were there more days with temperatures exceeding 90°F in 2018 or in 1948? For each of the 197 locations, they calculated the difference in number of days exceeding 90°F (number of days in 2018 – number of days in 1948). The average of these differences was 2.9 days with a standard deviation of 17.2 days.

1. Is there a relationship between the observations collected in 1948 and 2018? Or are the observations in the two groups independent? Explain. (2 points)

2. Write research and null hypotheses for this study in symbols and words. HINT: make sure to read the problem carefully. (2 points)

3. Find the critical value. Use α = .05. (2 points)

4. Calculate the test statistic. (3 points)

5. Make your decision about H₀ and write a sentence to interpret the result of the test. (3 points)

6. What type of error might you have made in this statistical test? Use the context of the problem to explain what this error would mean. (3 points)

7. Based on the results of this hypothesis test, would you expect a 95% confidence interval for the average difference between the number of days exceeding 90°F in 1948 and 2018 to include 0? Explain your reasoning(3 points)

Answer 1

Solution:

a)

Yes, the observations are paired on the basis o location

b)

Let D be the data of difference in no. of days. Then,

H0: $\mu _{D}$ = 0

H1:$\mu _{D}$ > 0

c)

The condition is that the sample of difference should belong to a normally distributed population. By looking at the histogram (as it looks like normal distribution) and since n>30. We can use normal approximation for this test.

d)

$t= \frac{(D-\mu _{D})}{\frac{s}{\sqrt{n}}}$

= (2.9-0)*$\sqrt{197}$ /17.2

= 2.9*14.036/17.2

= 2.366

df = n-1

= 196

p-value =  0.0095

e)

p-value =  0.0095 < 0.05 i.e. H0 can be rejected and hence we can say that there is enough evidence to conclude that there are more days with temperature exceeding 900F in 2018 than 1948.

f)

We could have made a type 1 error of 5% i.e. There is a 5% chance of there being no difference between the no. of days with temperature exceeding 900F in 2018 than 1948.

g)

No, It would not include 0. In fact, all the values will be positive in the confidence interval because we have rejected the null hypothesis which suggested the mean to be zero and confidence interval is for population mean which is not zero as per hypothesis test and hence confidence interval will not contain 'zero'

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