Is there enough evidence in the data that population average price of diamond for color “E”...

Is there enough evidence in the data that population average price of diamond for color “E” is more than 1500?

Solve in R using the t test.

Color	Price
D	1302
E	1510
G	1510
G	1260
D	1641
E	1555
F	1427
G	1427
H	1126
I	1126
F	1468
G	1202
E	1327
I	1098
E	1693
F	1551
G	1410
G	1269
H	1316
H	1222
E	1738
F	1593
G	1447
H	1255
F	1635
H	1485
F	1420
H	1420
F	1911
H	1525
F	1956
H	1747
I	1572
E	2942
G	2532
E	3501
E	3501
F	3501
F	3293
G	3016
F	3567
G	3205
D	3490
E	3635
F	3635
F	3418
D	3921
F	3701
F	3480
G	3407
E	3767
F	4066
E	4138
F	3605
G	3529
F	3667
I	2892
G	3651
G	3773
F	4291
E	5845
G	4401
G	4759
H	4300
F	5510
G	5122
H	5122
I	3861
F	5881
F	5586
F	5193
H	5193
F	5263
I	5441
I	4948
H	5705
F	6805
H	6882
H	6709
I	6682
E	3501
G	3432
F	3851
H	3605
E	3900
H	3415
H	4291
E	6512
E	5800
F	6285

Expert Solution

####Import the given file from Excel to R#####
>library(readxl)
>t_test <- read_excel("C:/Users/Admin/Desktop/t_test.xlsx")
>View(t_test)

####Now to extract the rows for E########
>install.packages("dplyr")
>library(dplyr)
>t=t_test %>% filter(Color=="E")

#######Performing t-test######
###Null Hypothesis: Mean=1200
###Alternate: Mean>1200
#### At 95% level of sognificance

>t.test(t$Price, mu=1500, alternative="greater",conf.level = 0.95)

Results from R is given as-

Based on the above results we have

t = 4.75, df = 15, p-value = 0.000129

since the p-value is less than 0.05(alpha value), therefore we have enough evidence to reject the null hypothesis that the average price of diamond for color “E” is more than 1500.

Thanks

milcah answered 2 weeks ago

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