Question

A group of transfer bound students wondered if they will spend the same mean amount on texts and supplies each year at their four-year university as they have at their community college. They conducted a random survey of 54 students at their community college and 66 students at their local four-year university. The sample means were $947 and $1,011, respectively. The population standard deviations are known to be $254 and $87, respectively. Conduct a hypothesis test to determine if the means are statistically the same assume alpha is 0.05.

Answer 1

Answer:

We accept H₀. We don´t have enough evidence to support the difference between the two means

Step-by-step explanation:

Group 1 : Comunity College

Sample:

size  :   n₁  = 54

Sample mean    x₁  = 947

Sample standard deviation     s₁ = 254

Group 2 : Local University

Sample

size  :  n₂  =  66

Sample mean :  x₂  = 1011

Sample standard deviation    s₂  =  87

Significance level  α  = 0,05      α  =  5 %      CI = 95 %

Both samples are with n> 30 so we can use z table for the test

from z-table we find  z(c)  =  1,96

The problem statement establishes to run a hypothesis test to determine if the means are statistically the same  ( or different) then we use two-tail-test

Hypothesis Test

Null Hypothesis                        H₀             x₁   -  x₂  = 0 or    x₁  = x₂

Alternative hypothesis             Hₐ             x₁  -  x₂  ≠  0  or    x₁ ≠ x₂

To calculate

z(s)  =  (  x₁   -  x₂  ) / √ s₁²/n₁  + s₂²/n₂ )

z(s)  =  947  -  1011 / √  (254)²/54   +  (87)² / 66

z(s) = -  64  /√ 1194,74  +  114,68

z(s) =  - 64  / 36,185

z(s) =  - 1,77

Comparing modules of  |z(c)|  and |z(s)|

z(s) < z(c)

Then z(s) is in the acceptance region we accept H₀ . There is not enough evidence to support differences in the means of the two groups