HELP ILL GIVE MOST BRAINLY AND 20 POINTS
1 answer:
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Answer: 8cm
Step-by-step explanation:
Solve for h;
Answer:
1 a
1b
1c
2a
2b
2c
Step-by-step explanation:
Considering question a
The alternative hypothesis is H1:μ>μ0
The test statistics is z =1.88
Generally from the z-table the probability of z =1.88 for a right tailed test is
Considering question b
The alternative hypothesis is H1:μ<μ0
The test statistics is z=−2.75
Generally from the z-table the probability of z=−2.75 for a left tailed test is
Considering question c
The alternative hypothesis is H1:μ≠μ0
The test statistics is z=2.88
Generally from the z-table the probability of z=2.88 for a right tailed test is
Generally the p-value for the two-tailed test is
=>
Considering question 2a
The alternative hypothesis is H1:μ>μ0
The sample size is n=16
The test statistic is t = 3.733
Generally the degree of freedom is mathematically represented as
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=>
Generally from the t distribution table the probability of t = 3.733 at a degree of freedom of for a right tailed test is
Considering question 2b
The alternative hypothesis is H1:μ<μ0
The degree of freedom is df=23
The test statistic is ,t= −2.500
Generally from the t distribution table the probability of t= −2.500 at a degree of freedom of df=23 for a left tailed test is
Considering question 2c
The alternative hypothesis is H1:μ≠μ0
The sample size is n= 7
The test statistic is ,t= −2.2500
Generally the degree of freedom is mathematically represented as
=>
=>
Generally from the t distribution table the probability of t= −2.2500 at a degree of freedom of for a left tailed test is
Generally the p-value for t= −2.2500 for a two tailed test is