Can you conclude with a high degree of confidence that the population means for Iowa and New Jersey students are different? A.
Because of the extremely low p-value, we cannot reject the null hypothesis with a very high degree of confidence. Hence, the population means for Iowa and New Jersey students are not different. B. Because of the extremely high p-value, we can reject the null hypothesis with a very high degree of confidence. Hence, the population means for Iowa and New Jersey students are different. C. Because of the extremely low p-value, we can reject the null hypothesis with a very high degree of confidence. Hence, the population means for Iowa and New Jersey students are different. D. Because of the extremely high p-value, we cannot reject the null hypothesis with a very high degree of confidence. Hence, the population means for Iowa and New Jersey students are not different.
1 answer:
Complete Question
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Answer:
Option A is correct
Step-by-step explanation:
From the question we are told that
The 95% confidence interval of the mean score of all New Jersey third graders is (58.957 , 63.043)
The sample size is n = 184
The sample average is
The standard deviation is
The 90% confidence interval for the difference in mean score between lowa and New Jersey is (-6.33 , -1.67 )
The p-value is
Generally given that the p-values is very low compared to the level of significance (100 -90 = 10% = 0.10 ) we can reject the null hypothesis which is usually a claim that the means of the population are the same )
Hence the population means are different .
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Answer:
And the explanation of this number is:"The number of text messages for Kendra it's 2.26 deviations above the mean"
Step-by-step explanation:
1) Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
2) Calculate the z score
Let X the random variable that represent the number of text messages per month, and for this case we know the distribution for X is given by:
Where and
And the best way to solve this problem is using the normal standard distribution and the z score given by:
If we apply this formula to our probability we got this:
And the explanation of this number is:"The number of text messages for Kendra it's 2.26 deviations above the mean"