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can anyone help me with that problem?

The school newspaper claims that 35% of students have credit card debt. You think it’s different. We simple random sample “M” students and “H” say that they have credit card debt. M=104 H= 30

Is there sufficient evidence at the 5% significance level (α = .05) to indicate that the percent of students with credit card debt is different than 35%

Hassan Othman
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    Welcome to MSE. You'll get a lot more help if you show that you have made a real effort to solve the problem yourself, even if you haven't made much progress. What are your thoughts? What have you tried? How far could you get? Where are you stuck? This question will likely be closed if you don't add more context. Please respond by editing the question body. Clarifications don't belong in the comments. – saulspatz Apr 18 '21 at 23:08
  • I suggest you do a one-proportion z-test: https://stats.libretexts.org/Bookshelves/Introductory_Statistics/Book%3A_Statistics_Using_Technology_(Kozak)/07%3A_One-Sample_Inference/7.02%3A_One-Sample_Proportion_Test – Stephen Donovan Apr 18 '21 at 23:10

1 Answers1

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Exact binomial test. You want to test $H_0: p = .35$ against $H_a: p\ne .36$ at 5% level. Your data are $x = 30$ successes in $n = 104$ trials. If $H_0$ is true then $X \sim \mathsf{Binom}(n=104, p=.35).$

In R statistical software, this test gives P-value about $0.2 > 0.05 = 5\%,$ so the null hypothesis is not rejected. One might say that the sample proportion $\hat p = 0.2885$ of successes is not significantly different from $0.35 = 35\%.$

binom.test(30, 104, p=.35)

    Exact binomial test




data:  30 and 104
number of successes = 30, number of trials = 104, p-value = 0.2173
alternative hypothesis: true probability of success is not equal to 0.35
95 percent confidence interval:
 0.2038192 0.3855334
sample estimates:
probability of success 
             0.2884615

Because you have shown no work of your own, there is no way for me to know whether this answer matches what you are studying in class. Maybe you are expected to use an approximate normal tests based on $Z = \frac{X - \mu}{\sigma},$ where $\mu = np_0 = 104(.35);$ $\sigma = \sqrt{np_0(1-p_0)},$ using standard normal CDF tables to provide an approximate P_value.

Minitab statistical software shows results of such an approximate test:

Test and CI for One Proportion

Test of p = 0.35 vs p ≠ 0.35


Sample   X    N  Sample p         95% CI         Z-Value  P-Value
1       30  104  0.288462  (0.201390, 0.375533)    -1.32    0.188


Using the normal approximation.

From R: Here is a plot of the $\mathsf{Binom}(104, .35)$ along with its approximating normal density curve.

enter image description here

If this response gives you some cue how to work the problem using what you have studied, perhaps you can edit your Question to show what you have been able to do.

Note: The sample size as well as the observed proportion of successes is important for this kind of test. If you had $90$ successes is $312$ trials, then there would be sufficient evidence to rejest $H_0.$

binom.test(90, 312, p=.35)

    Exact binomial test




data:  90 and 312
number of successes = 90, number of trials = 312, p-value = 0.02398
alternative hypothesis: true probability of success is not equal to 0.35
95 percent confidence interval:
  0.2388094 0.3421823
sample estimates:
probability of success 
             0.2884615
BruceET
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