Question

Each algorithm is applied to a group of 52 sample problems. The new algorithm completes the sample problems with a mean time of 24.04 hours. The current algorithm completes the sample problems with a mean time of 24.52 hours. The standard deviation is found to be 5.769 hours for the new algorithm, and 5.556 hours for the current algorithm. Conduct a hypothesis test at the 0.05 level of significance of the claim that the new algorithm has a lower mean completion time than the current algorithm.

Answer 1

Therefore we fail to reject the Null hypothesis
`H_0`

Because the is no sufficient evidence to support the claim of the alternative hypothesis
`H_a`

Explanation:

From the question we are told that

Sample size
`n_1 =52`

Mean time 1
`\=x_1=24.04\ hours`

Mean time 2
`\=x_2=24.52\ hours`

Standard deviation
`\partial _1=5.769`

Standard deviation
`\partial _2=5.556`

Level of significance 0.05

The null hypothesis is
`H_o: \mu_1=\mu_2`

Alternative hypothesis
`H_a: \mu_1<\mu_2`

Generally the test statistics is mathematically represented as

`Z= \frac{\=x_1-\=x_2}{\sqrt{(\sigma_1^2)/(n_1)+(\sigma_2^2)/(n_2)}}`

`Z= \frac{24.04-24.52}{\sqrt{(5.769^2)/(52)+(5.5562^2)/(52)}}`

`Z= -0.4321583379`

`Z= -0.43`

From Table the Z-critical value at
`\alpha =0.05\ is\ -1.6449`

Therefore

Zcal> Zcritical

(-0.43)=(-1.645)

Generally we fail to reject the null hypothesis
`H_0` at
`\alpha = 0.05`

Therefore we fail to reject the
`H_0`

Because the is no sufficient evidence to support the claim of
`H_a`