A single observation, $x$, is taken from a binomial distribution $B(20, p)$ and a value of 10 is obtained. Use this observation to test $H_0: p=0.3$ against $H_1: p>0.3$ using a 5% significance level.

A random variable has distribution $X \sim B(20, p)$. A single observation of $x=3$ is taken from this distribution. Test at the 1% significance level $H_0: p=0.45$ against $H_1: p<0.45$.

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04-11-2026
Mathematics

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A single observation, $x$, is taken from a binomial distribution $B(20, p)$ and a value of 10 is obtained. Use this observation to test $H_0: p=0.3$ against $H_1: p>0.3$ using a 5% significance level.

A random variable has distribution $X \sim B(20, p)$. A single observation of $x=3$ is taken from this distribution. Test at the 1% significance level $H_0: p=0.45$ against $H_1: p<0.45$.

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GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-04T07:28:35-05:00

For problem 3: The calculated p-value is $\approx 0.048$. Since $0.048 \le 0.05$, we reject the null hypothesis $H_0: p = 0.3$.
For problem 4: The calculated p-value is $\approx 0.0049$. Since $0.0049 \le 0.01$, we reject the null hypothesis $H_0: p = 0.45$.
Final conclusion: Both null hypotheses are rejected.
Problem 3: $\boxed{\text{Reject } H_0}$; Problem 4: $\boxed{\text{Reject } H_0}$

Explanation

Approach for Hypothesis Testing

To solve these hypothesis testing problems, we use the p-value method for a binomial distribution $X \sim B(n, p)$. The p-value represents the probability of observing the given result (or one more extreme in the direction of the alternative hypothesis $H_1$) assuming the null hypothesis $H_0$ is true. We compare this p-value to the chosen significance level $\alpha$. If the p-value $\le \alpha$, we reject $H_0$; otherwise, we fail to reject it.

Analysis of Problem 3

For problem 3, we have $X \sim B(20, p)$, with $H_0: p = 0.3$ and $H_1: p > 0.3$. The observed value is $x = 10$. The p-value is calculated as $P(X \ge 10 | p = 0.3)$.
The binomial probability is defined as $P(X = k) = \binom{20}{k} (0.3)^k (0.7)^{20-k}$.
Summing from $k = 10$ to $20$:
\[P(X \ge 10) = \sum_{k=10}^{20} \binom{20}{k} (0.3)^k (0.7)^{20-k} \approx 0.04796\]
Since $\alpha = 0.05$, and $0.04796 \le 0.05$, we reject the null hypothesis. There is sufficient evidence at the 5% level to suggest $p > 0.3$.

Analysis of Problem 4

For problem 4, we have $X \sim B(20, p)$, with $H_0: p = 0.45$ and $H_1: p < 0.45$. The observed value is $x = 3$. The p-value is calculated as $P(X \le 3 | p = 0.45)$.
Summing from $k = 0$ to $3$:
\[P(X \le 3) = \sum_{k=0}^{3} \binom{20}{k} (0.45)^k (0.55)^{20-k} \approx 0.00493\]
Since $\alpha = 0.01$, and $0.00493 \le 0.01$, we reject the null hypothesis. There is sufficient evidence at the 1% level to suggest $p < 0.45$.

Examples

These hypothesis tests are essential in quality control or clinical trials. For instance, if a manufacturer expects a defect rate of $p=0.3$ in a batch of 20 items and finds 10 defects, this analysis helps them decide if the process has worsened beyond acceptable standards, allowing them to take necessary corrective actions based on statistical confidence levels.