Question

The website Jane's Fighting Ships claims that at least 85% of the warships used in the Battle of Jutland in 1916 during the First World War were actual steel- hulled battleships. Find the Null Hypothesis, and the Alternative Hypothesis. (Do NOT perform the hypothesis test, just find the Null and Alternative hypotheses.)

Answer 1

Solution:

Question 1)

Given:

Claim: At least 85% of the warships used in the Battle of Jutland i 1916 during the First World War were actual steel-hulled battleships.

that is: $p \geq 0.85$

Direction of null and alternative hypothesis is depending on what we are asked in question.

Null hypothesis is always = type or or type.

Alternative hypothesis is either < or > or   type.

Thus null hypothesis is:

$H_{0}: p \geq 0.85$

and alternative hypothesis is:

Question 2)

Given:

Claim: the population proportion of people in California who have green eyes = 2%

Sample size = n = 650

x = Number of people in California who have green eyes = 17

thus sample proportion of people in California who have green eyes is:

$\hat{p}=\frac{x}{n} = \frac{17}{650} = 0.0262$

Level of significance = 5% = 0.05

Step 1) State H0 and H1:

$H_{0}: p = 0.02\: \: \: \: Vs \: \: \: H_{1}: p\neq 0.02$

H1 is not equal to type , since claim is non-directional and thus this is two tailed test.

Step 2) Test statistic:

$z= \frac{\hat{p}-p}{\sqrt{\frac{p\times (1-p)}{n}}}$

$z= \frac{0.0262 -0.02}{\sqrt{\frac{0.02\times (1-0.02)}{650}}}$

$z= \frac{ 0.0062 }{\sqrt{\frac{0.02\times 0.98}{650}}}$

$z= \frac{ 0.0062 }{\sqrt{ 0.0000301538 }}$

$z= \frac{ 0.0062 }{ 0.00549125 }$

Step 3) Find p-value:

For two tailed test , p-value is:

p-value = 2* P(Z > z test statistic) if z is positive

p-value = 2* P(Z < z test statistic) if z is negative

thus

p-value = 2* P(Z > z test statistic)

p-value = 2* P(Z > 1.12 )

p-value = 2* [ 1 - P(Z < 1.12 ) ]

Look in z table for z = 1.1 and 0.02 and find corresponding area.

17 .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 | 10.0 5000 15040 500 5120 .51605199 .5239 5279 .5319 .5359 | 10.1 .5398 .5438 .5478 .5517 .5557 5596 5636 .5675 15714 15753 |0.2 5793 5832 581 5910 5948 1987 .6026 6064 .6103 6141| 10.3 .6179 .6217 .6155 .6293 .6331 .6368 .6406 ,6443 .6480 ,6517 10.4 654 6591 .6286664,700.76 672,6808 ,6846879 10.5 6915 .6950 +85 7019 7054 1088 7123 17 7190 7224 | 0.6 72577291 124 1377389 7422 7454 7486 7517 7549 10. 7 7580 7611642 773 704 734 7647947823 (852 10.8 78817910 139 79677995 .8023 18051 .8078 .8106 .8133 10.9 .8159 .8186 12 18238 .8264 .8289 8315 18340 48365 8389 1108413.8438818485 850885318554 857.85998621 11.1 -869 86EE..8686 8708 18729 .8749 ,8770 .8790 18810 8830 ||

P( Z< 1.12) =0.8686

thus

p-value = 2* [ 1 - P(Z < 1.12 ) ]

p-value = 2* [ 1 - 0.8686 ]

p-value = 2* 0.1314

p-value = 0.2628

Step 4) Decision Rule:
Reject null hypothesis H0, if P-value < 0.05 level of significance, otherwise we fail to reject H0

Since p-value = 0.2628 > 0.05 level of significance, we fail to reject H0.

Step 5) Conclusion:

At 0.05 level of significance, we do not have sufficient evidence to reject the claim that the population proportion of people in California who have green eyes = 2% .

Thus we conclude that: the population proportion of people in California who have green eyes = 2% is true.