Let sample in Texas be p1, Louisiana be p2.
Null Hypothesis:
p1 - p2 = 0
Alternative Hypothesis: (This represents the claim)
p1- p2 < 0
To calculate test statistic, you need the pooled estimate which is a weighted average of the 2 sample proportions.
![P = \frac{n_1 p_1 + n_2 p_2}{n_1 + n_2} = \frac{2000(.179+.265)}{2000+2000} = 0.222](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7Bn_1%20p_1%20%2B%20n_2%20p_2%7D%7Bn_1%20%2B%20n_2%7D%20%3D%20%5Cfrac%7B2000%28.179%2B.265%29%7D%7B2000%2B2000%7D%20%3D%200.222)
Next get standard error:
![SE = \sqrt{P(1-P)(\frac{1}{n_1} + \frac{1}{n_2})} = \sqrt{.222(1-.222)(\frac{1}{2000}+\frac{1}{2000})} = 0.01314](https://tex.z-dn.net/?f=SE%20%3D%20%5Csqrt%7BP%281-P%29%28%5Cfrac%7B1%7D%7Bn_1%7D%20%2B%20%5Cfrac%7B1%7D%7Bn_2%7D%29%7D%20%3D%20%5Csqrt%7B.222%281-.222%29%28%5Cfrac%7B1%7D%7B2000%7D%2B%5Cfrac%7B1%7D%7B2000%7D%29%7D%20%3D%200.01314)
Calculate test statistic:
![Z = \frac{p_1 - p_2}{SE} = \frac{.179 - .256}{.01314} = -5.86](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7Bp_1%20-%20p_2%7D%7BSE%7D%20%3D%20%5Cfrac%7B.179%20-%20.256%7D%7B.01314%7D%20%3D%20-5.86)
To find p-value, look up Z-value in standard normal table.
Anything smaller than -3 or larger than 3, you can estimate to have
p-value = 0.
If p-value < alpha, Reject Null Hypothesis.
For this example, 0 < 0.05, therefore reject null hypothesis.
There is evidence to support claim that proportion of smokers in Texas is LESS than Louisiana.
Translation (up and down or side to side), reflection (flipping object over a line or point on plane), or dilation (making object bigger or smaller)
Answer:
idrk lol
Step-by-step explanation:
have a nice day tho