Question

Testing a claim about a population mean: t-Test

An overly involved "neighborhood watch" group has been investigating the length of lawns. In years past, lawns had been mowed to a mean length of 3.2 inches. A recent random sample of lawn lengths is given below. Conduct an appropriate test, at a 5% significance level, to determine if the overall mean lawn length is now higher than 3.2 inches.

Data (lawn lengths, measured in inches):

3.40
3.87
2.90
2.85
2.34
3.85
3.66
2.90
3.95
2.98
3.82
3.87
4.22
3.05
3.02
3.16
2.89
4.22

Checksum: 60.95

1. Carry out the procedure ("crunch the numbers"):

2. Sample mean: x¯≈
3. Standard deviation of the sample: s≈
4. Standard error of sample mean: σx¯≈   
5. Standardized test statistic: t≈
6. P-value:

Answer 1

Here in this scenario our claim is that the overall mean lown length is higher than 3.2 inches. To test this claim we have to use t distribution because here the population standard deviations is unknown.

Furth the one sample t test is performed at 0.05 level of significance as below,

Before first we have to calculate the sample mean and sample Standerd deviation as below,

The sample size is n = 18n=18. The provided sample data along with the data required to compute the sample mean \bar XXˉ and sample variance s^2s2 are shown in the table below:

	X	X2
	3.40	11.56
	3.87	14.9769
	2.90	8.41
	2.85	8.1225
	2.34	5.4756
	3.85	14.8225
	3.66	13.3956
	2.90	8.41
	3.95	15.6025
	2.98	8.8804
	3.82	14.5924
	3.87	14.9769
	4.22	17.8084
	3.05	9.3025
	3.02	9.1204
	3.16	9.9856
	2.89	8.3521
	4.22	17.8084
Sum =	60.95	211.603

The sample mean \bar XXˉ is computed as follows:

The t critical value is calculated using t table or using Excel at 17 degrees of freedom.

1. Carry out the procedure as given above.

2. Sample mean: x¯≈ 3.386
3. Standard deviation of the sample: s≈ 0.554
4. Standard error of sample mean: σx¯≈ s/√n = 0.554/√18 = 0.1306
5. Standardized test statistic: t≈ 1.424
6. P-value: 0.0862.

Thank you.