Question

A random sample of 23 chemists from Washington state shows an average salary of $41151 with a standard deviation of $775. A random sample of 17 chemists from Florida state shows an average salary of $46349 with a standard deviation of $924. A chemist that has worked in both states believes that chemists in Washington make a different amount than chemists in Florida. At αα=0.01 is this chemist correct? Let Washington be sample 1 and Florida be sample 2.

The correct hypotheses are:

• H0:μ1≤μ2H0:μ1≤μ2
  HA:μ1>μ2HA:μ1>μ2(claim)
• H0:μ1≥μ2H0:μ1≥μ2
  HA:μ1<μ2HA:μ1<μ2(claim)
• H0:μ1=μ2H0:μ1=μ2
  HA:μ1≠μ2HA:μ1≠μ2(claim)

Since the level of significance is 0.01 the critical value is 2.745 and -2.745

The test statistic is: (round to 3 places)

The p-value is: (round to 3 places)

The decision can be made to:

• reject H0H0
• do not reject H0H0

The final conclusion is that:

• There is enough evidence to reject the claim that chemists in Washington make a different amount than chemists in Florida.
• There is not enough evidence to reject the claim that chemists in Washington make a different amount than chemists in Florida.
• There is enough evidence to support the claim that chemists in Washington make a different amount than chemists in Florida.
• There is not enough evidence to support the claim that chemists in Washington make a different amount than chemists in Florida.

Answer 1

H0:μ1=μ2H0:μ1=μ2
HA:μ1≠μ2HA:μ1≠μ2(claim)

Test Statistic :-

t = ( 41151 - 46349 ) / \sqrt{ ( 775^{2} / 23) + ( 924^{2} / 17)}
t = -18.814

P - value = P ( t > 18.8136 ) = 0 < 0.01

Reject null hypothesis if P value < α = 0.01 level of significance
P - value = 0 < 0.01 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

There is enough evidence to support the claim that chemists in Washington make a different amount than chemists in Florida.