Researchers doing a study comparing time spent on social media and time spent on studying randomly sampled 200 students at a major university. They found that students in the sample spent an average of 2.3 hours per day on social media and an average of 18 hours per day on studying, If all the students at the university in fact spent 2.2 hours per day on studying, with a standard deviation of 2 hours, the sampling distribution of the time spent studying has approximate distribution:

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;

its the second choice.

step-by-step explanation:

the elements in the range will be 7 less than the elements in the domain.