Question

The following data from the Census bureau represents the total annual payroll from some California counties in 2015 5.8949 5.9822 6.4269 6.7192 6.7690 7.0776 7.2107 7.2501 7.8076 8.2101 and in 2016 5.9428 6.0060 6.4510 6.7530 6.8098 7.0913 7.25597.2820 7.8352 8.2427 The data has been transformed to make it look more normal. We want to test whether there was a statistically significant increase in payrolls from 2015 to 2016. (a) If we assume that the variances in both samples are the same, calculate the pooled (b) Calculate the appropriate t statistic for testing whether the mean is the same in 2015 as 0.05 level) estimate of the sample standard deviation in 2106. (c) What do you conclude? (Use an α

Answer 1

2 sample t test :

We can find the sample mean and standard deviation of the both years using excel function =AVERAGE( ) and =STDEV.S() respectively.

$\bar{x}$₁= 6.9348 , S₁ = 0.7382 n₁ = 10 and $\bar{x}$₂ = 6.967,S₂= 0.737 , n₂ = 10

a) Pooled estimate of standard deviation :

S = $\sqrt{\frac{(n_{1}-1)*s_{1}^{2}+(n_{2}-1)*s_{2}^{2}}{n_{1}+n_{2}-2}}$

S = $\sqrt{\frac{(9*0.7382^2)+(9*0.737^2)}{10+10-2}}$

S = 0.7376

b) H₀: µ₁ = µ₂    vs H_a: µ₁ ≠ µ₂   

t statistic :

t =   $\frac{\bar{x}_1 -\bar{x}_2 }{{s}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}$

t =   $\frac{6.9348 -6.967 }{0.7376*\sqrt{\frac{1}{10}+\frac{1}{10}}}$  

t = -0.09762

c) We are given α =0.05,

Test statistic follows t distribution with degrees of freedom (df) = n₁+ n₂ - 2 = 10+10-2 = 18

We can find critical value using excel function =TINV( α , d.f )

=TINV( 0.05 , 18 ) = 2.1009

Critical value = 2.1009

Critical region: Reject H₀, if | t | ≥ 2.1009 Or fail to reject H0 , if |t| < 2.1009

|t |is absolute value of t statistic.

Decision : we have t = -0.09762 , so |t| = 0.09762

As |t| < 2.1009 we fail to reject H₀

Conclusion : We do not have significant evidence that mean in year 2015 is different than mean in year 2016