Answer:
The 98% confidence interval for the true proportion of women shoppers who shop on impulse is
[tex]0.1391 < p < 0.2009[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 800
The sample proportion is [tex]\r p = 0.17[/tex]
Given that the confidence level is 98%
The level of significance is evaluated as
[tex]\alpha = 100 -98[/tex]
[tex]\alpha = 2[/tex]%
[tex]\alpha = 0.02[/tex]
given that this is a two tailed test
[tex]\frac{\alpha }{2} = \frac{0.02}{2} = 0.01[/tex]
The critical values obtained from the normal distribution table is
[tex]z_{\frac{\alpha }{2} } = 2.33[/tex]
Now the the margin of error is mathematically evaluated as
[tex]MOE = 2.33 * \sqrt{\frac{0.17 (1-0.17)}{800} }[/tex]
[tex]MOE = 0.0309[/tex]
the 98% confidence interval for the true proportion of women shoppers who shop on impulse is mathematically evaluated as
[tex]0.17 - 0.0309 < p < 0.17 + 0.0309[/tex]
[tex]0.1391 < p < 0.2009[/tex]
The 98% confidence interval for the true proportion of women shoppers will be:
"0.1391 < p < 0.2009".
According to the question,
Sample size, n = 800
Sample proportion, [tex]\hat p[/tex] = 0.17
Confidence level = 98%
Now,
The level of significance will be;
→ α = 100 - 98
= 2% or,
= 0.02
Two-tailed be:
→ [tex]\frac{\alpha}{2}[/tex] = [tex]\frac{0.02}{2}[/tex]
= 0.01
The critical value be:
[tex]z_{\frac{\alpha}{2} }[/tex] = 2.33
then, The margin of error be:
= 2.33 × [tex]\sqrt{\frac{0.17(1-0.17)}{800} }[/tex]
= 2.33 × [tex]\sqrt{\frac{0.17\times 0.83}{800} }[/tex]
= 2.33 × [tex]\sqrt{\frac{0.1411}{800} }[/tex]
= 0.0309
hence,
The 98% confidence level be:
= 0.17 - 0.0309 < p < 0.17 < 0.0309
= 0.1391 < p < 0.2009
Thus the above approach is correct.
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https://brainly.com/question/16112320
