Question

Suppose that the neighboring cities of Tweed and Ledee are long-term rivals. Neal, who was born and raised in Tweed, is confident that Tweed residents are more concerned about the environment than the residents of Ledee. He knows that the average electricity consumption of Tweed households last February was 854.11 kWh and decides to test if Ledee residents used more electricity that month, on average. He collects data from 65 Ledee households and calculates the average electricity consumption to be 879.28 kWh with a standard deviation of 133.29 kWh. There are no outliers in his sample data. Neal does not know the population standard deviation nor the population distribution. He uses a one-sample t-test with a significance level of α = 0.05 to test the null hypothesis, H0:µ=854.11, against the alternative hypothesis, H1:μ>854.11 , where μ is the average electricity consumption of Ledee households last February. Neal calculates a t‑statistic of 1.522 and a P-value of 0.066.

Based on these results, complete the following sentences to state the decision and conclusion of the test.

Neal's decision is to__________ the __________ (p 0.066). There is_________ evidence to _________ the claim that the average electricity consumption of ____________ is _________ , ________

Answer 1

Complete Question

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Answer:

Neal's decision is to fail to reject the null hypothesis   (p 0.066). There is no sufficient evidence to prove the claim that the average electricity consumption of all  Ledee household is greater than , 854.28 kWh

Step-by-step explanation:

From the question we are told that

   The population mean is  $\mu = 854.11$

   The sample size is  $n = 65$

    The sample mean is  $\= x = 879.28 \ kWh$

    The standard deviation is  $\sigma = 133.29 \ kWh$

    The level of significance is  $\alpha = 0.05$

     The  null hypothesis is  $H_o: \mu = 854.11$

     The  alternative hypothesis is  $H_a : \mu > 854.11$

     The  t-statistics is  $t = 1.522$

      The  p-value is $p-value = 0.066$

Now from the given data we can see that

         $p-value < \alpha$

Generally when this is the case , we fail to reject the null hypothesis

   So

Neal's decision is to fail to reject the null hypothesis   (p 0.066). There is no sufficient evidence to prove the claim that the average electricity consumption of all  Ledee household is greater than , 854.28 kWh