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I'm currently taking a statistics course and am a little confused about this question. I've found true proportions before, but usually I'm given numbers, rather than just percentages. So, I'm very confused about the following question. Any help and/or guidance would be very much appreciated:

A survey of $1428$ American adults indicates that $52.9\%$ of phone owners browse the Internet on their phone rather than on a computer, up from $49.7\%$ in the previous year.

At a significance level of $α = 5\%$, test whether the true proportion of adult phone owners who browse on their phone increases from $49.7\%$ in the previous year.

How would I go about solving this? I'm not sure if it's relevant to this problem, but I've found the $95\%$ Wald's confidence interval, which is:

$0.529 + 1.96 \cdot \sqrt{ \frac{0.529 \cdot (1 - 0.529)}{1428}}$

$0.529 - 1.96 \cdot \sqrt{ \frac{0.529 \cdot (1 - 0.529)}{1428}}$

$[0.5031101, 0.5548899]$

Other than doing this, I have no idea how to go about solving this. Also, for this course we use RStudio, so any answers that include R are more than welcome!

vitamin d
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larn
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    Are you familiar with how to find a z-score using sample proportion? – c_gnar Mar 26 '21 at 19:29
  • @c_gnar Vaguely? We've discussed it once or twice. Would I find the z-score by doing this: (0.529 - 0.497) / sqrt( (0.497 * (1 - 0.497)) / 1428) ? If that's incorrect then, no, I'm not very familiar with finding a z-score using sample proportion :') – larn Mar 26 '21 at 19:32
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    Making a two-sided confidence interval is a reasonable thing to try. (+`) However, I think you want a one-sided test (because of the word "increases" in the question. Also, your binomial CI is not found by "inverting" a binomial test, so the fact that a CI does not cover the hypothetical value is not precisely a criterion for rejection. – BruceET Mar 27 '21 at 02:08

1 Answers1

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Test $H_0: p = 0.497$ against $H_a: p > 0.497$ at 5% level.

Using normal approximation, which seems OK with success probabilities near to $1/2$ and $n > 1000,$ one gets $$Z = \frac{\hat p - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} = \frac{.529-.497}{\sqrt{\frac{.497(1-.497)}{1428}}} = 2.419 > 1.645,$$ so reject $H_0$ at the 5% level of significance. [Using R for computations.]

p.est = 0.529;  p.0 = 0.497;  n = 1428
se = sqrt(p.0*(1-p.0)/n)
z = (p.est-p.0)/se; z
[1] 2.418532

qnorm(.95)
[1] 1.644854


1 - pnorm(z)
[1] 0.007791631

Regrettably, $n\hat p = 755.412,$ using values stated in the problem, is not an integer. I used $755$ successes in Minitab, which gives the following output--in substantial agreement with the above.

Test and CI for One Proportion

Test of p = 0.497 vs p > 0.497


Sample    X     N  Sample p  95% Lower Bound  Z-Value  P-Value
1       755  1428  0.528711         0.506984     2.40    0.008

Using the normal approximation.

An exact binomial test, have the number of successes $X \sim \mathsf{Binom}(n = 1428, p_0 = 0.497).$ Thus the critical value $c$ such that $P(X \ge c\,|\, H_0) \approx 0.05,$ without exceeding $0.05.$ So the critical value is $c = 716$ and we reject $H_0$ because the observed number of successes $X = 755 > c = 716.$ Also, the P-value is $P(X \ge 755\,|\, H_0)$ $= 1 - P(X \le 754\, |\, H_0)$ $= 0.009 < 0.05 = 5\%.$

qbinom(.95, 1428, 0.479)
[1] 715
1 - pbinom(715, 1428, 0.479)
[1] 0.04770684  # close to 5% without exceeding
1 - pbinom(716, 1428, 0.479)
[1] 0.04267659
1 - pbinom(714, 1428, 0.479)
[1] 0.05320117

1 - pbinom(754, 1428, 0.497)
[1] 0.008878225
BruceET
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