document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1169008>Question 1169008</A>: <I><SPAN STYLE=\"word-break: break-all;\">  A group of students estimated the length of one minute without reference to a watch or​ clock, and the times​ (seconds) are listed below. Use a 0.10 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one​ minute? <BR>\n" );
document.write( "70 84 42 65 41 24 58 63 69 49 61 72 90 91 67<BR>\n" );
document.write( "Need to find<BR>\n" );
document.write( "Hypothesis test <BR>\n" );
document.write( "p-value  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851459 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=851459\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to conduct the hypothesis test using the given data and significance level:\r<BR>\n" );
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document.write( "**1. Define the Hypotheses:**\r<BR>\n" );
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document.write( "* **Null Hypothesis ($H_0$):** The population mean of the estimated times is equal to 60 seconds.<BR>\n" );
document.write( "    $H_0: \mu = 60$<BR>\n" );
document.write( "* **Alternative Hypothesis ($H_a$):** The population mean of the estimated times is not equal to 60 seconds. This is a two-tailed test because we are testing if the mean is *different* from 60 seconds.<BR>\n" );
document.write( "    $H_a: \mu \neq 60$\r<BR>\n" );
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document.write( "**2. Set the Significance Level:**\r<BR>\n" );
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document.write( "The significance level is given as $\alpha = 0.10$.\r<BR>\n" );
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document.write( "**3. Calculate the Sample Statistics:**\r<BR>\n" );
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document.write( "First, we need to calculate the sample mean ($\bar{x}$) and the sample standard deviation ($s$) from the given data:\r<BR>\n" );
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document.write( "Data: 70, 84, 42, 65, 41, 24, 58, 63, 69, 49, 61, 72, 90, 91, 67\r<BR>\n" );
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document.write( "* **Sample Size ($n$) = 15**\r<BR>\n" );
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document.write( "* **Calculate the sample mean ($\bar{x}$):**<BR>\n" );
document.write( "    $\bar{x} = \frac{70 + 84 + 42 + 65 + 41 + 24 + 58 + 63 + 69 + 49 + 61 + 72 + 90 + 91 + 67}{15}$<BR>\n" );
document.write( "    $\bar{x} = \frac{946}{15} \approx 63.07$ seconds\r<BR>\n" );
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document.write( "* **Calculate the sample standard deviation ($s$):**<BR>\n" );
document.write( "    To do this, we first find the deviations from the mean, square them, sum the squared deviations, divide by $n-1$, and then take the square root.\r<BR>\n" );
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document.write( "    | Value | Deviation ($x - \bar{x}$) | Squared Deviation ($(x - \bar{x})^2$) |<BR>\n" );
document.write( "    |-------|--------------------------|--------------------------------------|<BR>\n" );
document.write( "    | 70    | $70 - 63.07 = 6.93$     | $6.93^2 = 48.0249$                   |<BR>\n" );
document.write( "    | 84    | $84 - 63.07 = 20.93$    | $20.93^2 = 438.0649$                  |<BR>\n" );
document.write( "    | 42    | $42 - 63.07 = -21.07$   | $(-21.07)^2 = 443.9449$                 |<BR>\n" );
document.write( "    | 65    | $65 - 63.07 = 1.93$     | $1.93^2 = 3.7249$                    |<BR>\n" );
document.write( "    | 41    | $41 - 63.07 = -22.07$   | $(-22.07)^2 = 487.0849$                 |<BR>\n" );
document.write( "    | 24    | $24 - 63.07 = -39.07$   | $(-39.07)^2 = 1526.4649$                |<BR>\n" );
document.write( "    | 58    | $58 - 63.07 = -5.07$    | $(-5.07)^2 = 25.7049$                   |<BR>\n" );
document.write( "    | 63    | $63 - 63.07 = -0.07$    | $(-0.07)^2 = 0.0049$                    |<BR>\n" );
document.write( "    | 69    | $69 - 63.07 = 5.93$     | $5.93^2 = 35.1649$                    |<BR>\n" );
document.write( "    | 49    | $49 - 63.07 = -14.07$   | $(-14.07)^2 = 197.9649$                 |<BR>\n" );
document.write( "    | 61    | $61 - 63.07 = -2.07$    | $(-2.07)^2 = 4.2849$                    |<BR>\n" );
document.write( "    | 72    | $72 - 63.07 = 8.93$     | $8.93^2 = 79.7449$                    |<BR>\n" );
document.write( "    | 90    | $90 - 63.07 = 26.93$    | $26.93^2 = 725.2249$                  |<BR>\n" );
document.write( "    | 91    | $91 - 63.07 = 27.93$    | $27.93^2 = 780.0849$                  |<BR>\n" );
document.write( "    | 67    | $67 - 63.07 = 3.93$     | $3.93^2 = 15.4449$                    |<BR>\n" );
document.write( "    | **Sum** |                       | **4820.0000** |\r<BR>\n" );
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document.write( "    $s^2 = \frac{\sum(x - \bar{x})^2}{n-1} = \frac{4820}{15 - 1} = \frac{4820}{14} \approx 344.2857$<BR>\n" );
document.write( "    $s = \sqrt{344.2857} \approx 18.56$ seconds\r<BR>\n" );
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document.write( "**4. Determine the Test Statistic:**\r<BR>\n" );
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document.write( "Since the population standard deviation is unknown and the sample size is small ($n < 30$), we will use a t-test. The test statistic is:\r<BR>\n" );
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document.write( "$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$\r<BR>\n" );
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document.write( "where:<BR>\n" );
document.write( "* $\bar{x}$ is the sample mean ($\approx 63.07$)<BR>\n" );
document.write( "* $\mu_0$ is the hypothesized population mean ($60$)<BR>\n" );
document.write( "* $s$ is the sample standard deviation ($\approx 18.56$)<BR>\n" );
document.write( "* $n$ is the sample size ($15$)\r<BR>\n" );
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document.write( "$t = \frac{63.07 - 60}{18.56 / \sqrt{15}}$<BR>\n" );
document.write( "$t = \frac{3.07}{18.56 / 3.873}$<BR>\n" );
document.write( "$t = \frac{3.07}{4.792} \approx 0.641$\r<BR>\n" );
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document.write( "**5. Determine the P-value:**\r<BR>\n" );
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document.write( "For a two-tailed t-test with $n-1 = 15 - 1 = 14$ degrees of freedom, we need to find the p-value associated with a test statistic of $t \approx 0.641$.\r<BR>\n" );
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document.write( "Looking at a t-distribution table or using a statistical calculator, the p-value for a two-tailed test with $t = 0.641$ and 14 degrees of freedom is greater than $2 \times 0.25 = 0.50$. More precisely, it's approximately $0.532$.\r<BR>\n" );
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document.write( "**Hypothesis Test:**\r<BR>\n" );
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document.write( "* **Test Statistic:** $t \approx 0.641$<BR>\n" );
document.write( "* **Degrees of Freedom:** $df = 14$<BR>\n" );
document.write( "* **Type of Test:** Two-tailed\r<BR>\n" );
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document.write( "**P-value:**\r<BR>\n" );
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document.write( "* **P-value $\approx 0.532$**\r<BR>\n" );
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document.write( "**Conclusion:**\r<BR>\n" );
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document.write( "We compare the p-value to the significance level ($\alpha = 0.10$).\r<BR>\n" );
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document.write( "Since the p-value ($0.532$) is greater than the significance level ($0.10$), we **fail to reject the null hypothesis**.\r<BR>\n" );
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document.write( "**Answer to the question: Does it appear that students are reasonably good at estimating one minute?**\r<BR>\n" );
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document.write( "Based on this hypothesis test at a 0.10 significance level, there is **not enough statistical evidence** to conclude that the mean estimated time by students is significantly different from 60 seconds. Therefore, it appears that students are **reasonably good** at estimating one minute, as the data does not provide strong evidence to suggest their estimates are consistently off.\r<BR>\n" );
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document.write( "Final Answer: The final answer is $\boxed{p-value \approx 0.532}$ </SPAN>\n" );
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