Question

Can you please show both in RStudio code? Thank You

Airports:

The temperature is recorded at 60 airports in a region. The average temperature is 68 degrees Fahrenheit with a standard deviation of 5 degrees. The last known average temperature from all airports is 67 degrees Fahrenheit.   Is the recorded temperature at the 60 airports different from the average temperature at all airports?

• Answer by determining statistical significance using the test statistic, degrees of freedom, and p-value.
• Answer substantive significance with difference in means and effect size.

New York Attitudes:

The New York Chamber of Commerce has asked you to do a study concerning people's attitudes toward our city. As part of the study, you will ask them to rate their image of New York on a scale from 1 to 100 (1= awful city - call in the bulldozers; 100 = wonderful city - there is none better). Previous data show this scale is normally distributed with a last known population average rating of 50 and a standard deviation of 10.

Answer 1

Q1. ANSWERING FIRST QUESTION.

Given that,
population mean(u)=67
sample mean, x =68
standard deviation, s =5
number (n)=60
null, Ho: μ=67
alternate, H1: μ!=67
level of significance, alpha = 0.05
from standard normal table, two tailed t alpha/2 =2.001
since our test is two-tailed
reject Ho, if to < -2.001 OR if to > 2.001
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =68-67/(5/sqrt(60))
to =1.5492
| to | =1.5492
critical value
the value of |t alpha| with n-1 = 59 d.f is 2.001
we got |to| =1.5492 & | t alpha | =2.001
make decision
hence value of |to | < | t alpha | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 1.5492 ) = 0.1267
hence value of p0.05 < 0.1267,here we do not reject Ho
ANSWERS
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null, Ho: μ=67
alternate, H1: μ!=67
test statistic: 1.5492
critical value: -2.001 , 2.001
decision: do not reject Ho
p-value: 0.1267