9.7 Hypothesis testing: two column model step by step example of

• Is the problem about numerical or categorical data?
• If the data is numerical is the population standard deviation known?
• Do you have one group or two groups?
• What type of model do we have?

1. Express the hypothesis to be tested in symbolic form,
2. Write a symbolic expression that must be true when the original claim is false.
3. The null hypothesis is the statement which includes the equality.
4. The alternative hypothesis is the statement without the equality.

Null hypothesis in words: The null hypothesis is that the true mean height of the loaves is equal to 15 cm.

Null Hypothesis symbolically: $H_{\mathrm{o\; (Mean\; height)}}$ : μ = 15

Alternative Hypothesis in words: The alternative is that the true mean height on average is greater than 15 cm.

Alternative Hypothesis symbolically: $H_{\mathrm{a\; (Mean\; height)}}$ : μ>15

Randomization Condition: The sample is a random sample.
Independence Assumption: It is reasonable to think that the loaves of bread have heights that are independent.
10% Condition: I assume the number of loaves of bread baked is more than 100, so 10 loaves is less than 10% of the population.
Sample Size Condition: Since the distribution of the bread heights is normal, my sample of 10 loaves is large enough.

Determine the P-value

State whether you reject or fail to reject the Null hypothesis.

If you reject the null hypothesis, continue to complete the following

Calculate and display your confidence interval for the Alternative hypothesis.

$\overline{x}=\mathrm{15.7};\sigma =1;n=\mathrm{10};\mathrm{SE}=\left(\frac{\sigma }{\sqrt{n}}\right)=\frac{1}{\sqrt{\mathrm{10}}}=\mathrm{0.3162}$

z-score for a two tailed test with 95% confidence is plus or minus 1.96 (read from the z-table 0.025 probability in the left tail and 0.025 probability in the right tail)

• $\overline{x}-\mathrm{z*}(SE)\mathrm{<\mu <}\overline{x}+\mathrm{z*}(SE)$
• $\overline{x}-1.96(SE)\mathrm{<\mu <}\overline{x}+1.96(SE)$ ;

$15.7-1.96\mathrm{(0.3162)}\mathrm{<\mu <}15.7+1.96\mathrm{(0.3162)}$ ;

$15.08\mathrm{<\mu <}16.32$

Conclusion State your conclusion based on your confidence interval.