A public health researcher wishes to study the dietary behavior of residents in Durham County. The researcher randomly contacts
35 county residents and collects data on their daily sugar intake and obtained a sample average of 37.4 grams sugar per day and a sample standard deviation of 4.2 grams per day. The researcher would like to construct a 99% confidence interval for the mean daily sugar intake of residents in the county using the data. Which distribution should the researcher use when analyzing these data?a. The researcher should use a normal distribution because the data come from a simple random sample and the sample is large.b. The researcher should use a normal distribution because the true population average daily sugar intake is likely not near zero.c. The researcher should use a normal distribution because they do not know the true standard deviation of daily sugar intake in Durham county.d. The researcher should use a t-distribution because the sample size is too small to accurately calculate a 99% confidence interval using a normal distribution.e. The researcher should use a t-distribution because they do not know the true standard deviation of daily sugar intake in Durham county.f. The researcher cannot determine whether a t-distribution or a normal distribution would be more appropriate without looking at the shape of the population distribution.
(e) The researcher should use a t-distribution because they do not know the true standard deviation of daily sugar intake in Durham county.
Step-by-step explanation:
A t-distribution is used when the true standard deviation (population standard deviation) is unknown. t-distribution uses the standard deviation of the samples.
In this case, it is the sample standard deviation that is known, therefore, the researcher should use a t-distribution.
The researcher should use a normal distribution because the data come from a simple random sample and the sample is large.
Step-by-step explanation:
The research should use a normal distribution because a simple random distribution was used to get the sample and the sample is large(> = 30).
With the parameters given standard deviation, the sample size, and the population mean, can be used to find the standard score. Which signifies that a normal distribution is the best approach to use.
5^2 -7•4+(36-2^(5) so do (36-2^5 first put that in the calculator which gets u 32.Then bring down 5^2-7•4+32 then do 5^2 and 7×4 which 25 -28+32 add them together u get 60 so 25 -60 is 35
