Question

The average temperature in St. Louis in November of 2018 was 38.8oF38.8oF . The unusually cold weather in St. Louis and other places around the country prompted President Trump to tweet at the time that this was proof that global warming cannot be real. Let’s put that claim to a statistical test:

According to the National Oceanic and Atmospheric Administration (NOAA, retrieved from www.weather.gov), the average daily temperature in St. Louis in November (starting in 1838) is 45.05oF use this link for data https://docs.google.com/spreadsheets/d/e/2PACX-1vS1p0V616KaIEvftqRaAUb2X1xbhc5todiGqid5YXkhHOSd8lb19cts7JfxvkTQkqSKHMdyuxy4eyVx/pubhtml?gid=1366414992&single=true

Look at the data over the 15-year span from 2004 to 2018Opens externally to conduct a hypothesis test about whether or not the average temperature in November in St. Louis is different than its historical average of 45.05oF45.05oF . Use 0.1 as your significance level.

1. The null and alternative hypothesis would be:
   • H0:μ=45.05
     H1:μ>45.05
   • H0:μ=45.05
     H1:p<45.05
   • H0:μ=45.05
     H1:μ≠45.05
   • H0:μ=45.05
     H1:μ<45.05
   • H0:p=45.05
     H1:p≠45.05
   • H0:p=45.05
     H1:p>45.05
2. The test is:

   left-tailed

   right-tailed

   two-tailed

3. The P-value of this sample is  (to 4 decimals).
4. Based on this we: Select an answer Fail to reject the null hypothesis Reject the null hypothesis
5. We address the original claim with the conclusion
   • There is insufficient evidence to support the claim that the average daily temperature is 45.05 with 0.1 significance level.
   • The average daily temperature in November is not 45.05 with 0.1 significance level.
   • The average daily temperature in November is 45.05 with 0.1 significance level.
   • There is sufficient evidence to support the claim that the average daily temperature is 45.05 with 0.1 significance level.
6. Does the conclusion in (e) mean that global warming is real? Please explain your answer.

Answer 1

The null and alternative hypothesis would be

H0:μ=45.05
H1:μ≠45.05

The test is:

two-tailed

The t-statistic is computed as follows: X - μo t= = 45.04560 – 45.05 3.9518/182 = -0.015 8/ √n (4) Decision about the null hypothesis Since it is observed that t = 0.015 Ste = 1.653, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p= 0.988, and since p= 0.988 > 0.10, it is concluded that the null hypothesis is not rejected.

Conclusion :

There is sufficient evidence to support the claim that the average daily temperature is 45.05 with 0.1 significance level.