Question

Refer to the accompanying data set and construct a 90​% confidence interval estimate of the mean pulse rate of adult​ females; then do the same for adult males. Compare the results. LOADING... Click the icon to view the pulse rates for adult females and adult males. Construct a 90​% confidence interval of the mean pulse rate for adult females. nothing bpmless thanmuless than nothing bpm ​(Round to one decimal place as​ needed.) Construct a 90​% confidence interval of the mean pulse rate for adult males. nothing bpmless thanmuless than nothing bpm ​(Round to one decimal place as​ needed.)

Compare the results. A. The confidence intervals do not​ overlap, so it appears that there is no significant difference in mean pulse rates between adult females and adult males. B. The confidence intervals do not​ overlap, so it appears that adult females have a significantly higher mean pulse rate than adult males. C. The confidence intervals​ overlap, so it appears that there is no significant difference in mean pulse rates between adult females and adult males. D. The confidence intervals​ overlap, so it appears that adult males have a significantly higher mean pulse rate than adult females.

Restaurant X   Restaurant Y
87   96
121   130
116   152
150   117
265   180
180   132
126   106
151   128
167   124
217   129
337   133
307   133
176   228
117   212
154   293
147   124
96   89
234   134
235   241
186   139
155   147
199   197
164   150
117   148
67   132
197   147
181   153
115   130
145   168
172   133
192   238
200   231
234   255
191   239
354   231
310   173
208   87
196   109
179   53
191   172
102   79
144   149
174   144
151   99
172   129
154   152
166   127
125   185
140   146
305   127

Answer 1

TRADITIONAL METHOD
given that,
sample mean, x =179.38
standard deviation, s =62.976
sample size, n =50
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 62.976/ sqrt ( 50) )
= 8.906
II.
margin of error = t α/2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 49 d.f is 1.677
margin of error = 1.677 * 8.906
= 14.936
III.
CI = x ± margin of error
confidence interval = [ 179.38 ± 14.936 ]
= [ 164.444 , 194.316 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =179.38
standard deviation, s =62.976
sample size, n =50
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 49 d.f is 1.677
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 179.38 ± t a/2 ( 62.976/ Sqrt ( 50) ]
= [ 179.38-(1.677 * 8.906) , 179.38+(1.677 * 8.906) ]
= [ 164.444 , 194.316 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 164.444 , 194.316 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean

b.
TRADITIONAL METHOD
given that,
sample mean, x =153
standard deviation, s =50.023
sample size, n =50
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 50.023/ sqrt ( 50) )
= 7.074
II.
margin of error = t α/2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 49 d.f is 1.677
margin of error = 1.677 * 7.074
= 11.864
III.
CI = x ± margin of error
confidence interval = [ 153 ± 11.864 ]
= [ 141.136 , 164.864 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =153
standard deviation, s =50.023
sample size, n =50
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 49 d.f is 1.677
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 153 ± t a/2 ( 50.023/ Sqrt ( 50) ]
= [ 153-(1.677 * 7.074) , 153+(1.677 * 7.074) ]
= [ 141.136 , 164.864 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 141.136 , 164.864 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean

Answer:
a.
we are 90% sure that the interval [ 164.444 , 194.316 ] for females
b.
we are 90% sure that the interval [ 141.136 , 164.864 ] for males
option:B.
The confidence intervals do not​ overlap,
so it appears that adult females have a significantly higher mean pulse rate than adult males.