Question

A team of health research wants to investigate whether having regular lunch break improves working adults’ sleep. A group of 85 adults with a full-time job were recruited in the study, they reported average hours of sleep in the past week, committed to having a 1-hour, out-of-office, work-free lunch break each day for 3 months, and, at the end of the study, reported average hours of sleep in the past week. The mean difference in hours of sleep before and after the study was 3 with a standard deviation of 0.9. Which one of the following statements is INCORRECT? (Set alpha level at 0.05.)

The decision should be to reject the null hypothesis.

The null hypothesis is that hours of sleep remain the same before and after the study.

The obtained test statistic is 3.073.

The degrees of freedom is 84.

Answer 1

answer:

given data,

the group of adults is 85

the  committed to having to the 1-hour, out of office the is  work free for the lunch break of the each day is  for the 3 months,

The mean is 3

the standard deviation is as 0.9.

to find the one of the following statement is correct is....

the probability is shown in the below,

here we are know as the value is already,

we are know as the formula is the test statistic z is the below,

committed time /probability of the 3 months as mean / sqre n

now substitute the values is in the above formula,

   1/ ( 0.3 / sqre 85 )

       1 / ( 0.3 /9.219544 )

   1 / ( 0.03253957 )

   30.7318136

   30.73

note here the test statistic has as the not the 3.073 is that the 30.73

so finally the incorrect statement is as the option c

so finally the correct option is ( c )