Question

1. Use the diamond data:

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

Please solve using R please.

Thank you

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

Answer 1

Soln

data_5Nov = read.csv(file.choose(),header = T)

head(data_5Nov)

Output

Color Price

1     D 1302

2     E 1510

3     G 1510

4     G 1260

5     D 1641

6     E 1555

data = data_5Nov[data_5Nov$Color=="E",] ## to get only the ‘E’ Color records

Alpha = 0.05

Null and Alternate Hypothesis

H₀: µ = 1500

H_a: µ > 1500

t.test(data$Price, mu = 1500,alternative = "greater")
Output
        One Sample t-test

data:  data$Price

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

alternative hypothesis: true mean is greater than 1500

95 percent confidence interval:

 2717.118      Inf

sample estimates:

mean of x 

 3429.062 

Result

Since the p-value is less than 0.05, we reject the null hypothesis in favour of alternate hypothesis.

Conclusion

Population average price of diamond for color “E” is more than 1500.