document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1181347>Question 1181347</A>: <I><SPAN STYLE=\"word-break: break-all;\"> According to a survey, the typical American spends 154.8 minutes (2.58 hours) per day watching television. A survey of 50 Internet users results in a mean time watching television per day of 128.7 minutes per day with a standard deviation of 46.5 minutes. Conduct the appropriate test to determine if Internet users spend less time watching television than the average American at α = 0.05 level of significance.\r<BR>\n" );
document.write( "\n" );
document.write( "[Remember, you should include your H0 , H1 , your p-value, and a conclusion statement explaining whether you accept or reject the claim.]<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850059 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=850059\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to conduct a hypothesis test for this scenario:\r<BR>\n" );
document.write( "\n" );
document.write( "**1. State the Hypotheses:**\r<BR>\n" );
document.write( "\n" );
document.write( "*   **Null Hypothesis (H0):** Internet users spend the same amount of time watching television as the average American (μ = 154.8 minutes).<BR>\n" );
document.write( "*   **Alternative Hypothesis (H1):** Internet users spend less time watching television than the average American (μ < 154.8 minutes).  This is a one-tailed test (left-tailed).\r<BR>\n" );
document.write( "\n" );
document.write( "**2. Significance Level:** α = 0.05\r<BR>\n" );
document.write( "\n" );
document.write( "**3. Test Statistic:**\r<BR>\n" );
document.write( "\n" );
document.write( "Since we have a sample size greater than 30, we can use a z-test. The formula for the z-statistic is:\r<BR>\n" );
document.write( "\n" );
document.write( "z = (sample mean - population mean) / (standard deviation / √sample size)\r<BR>\n" );
document.write( "\n" );
document.write( "z = (128.7 - 154.8) / (46.5 / √50)<BR>\n" );
document.write( "z = -26.1 / (46.5 / 7.07)<BR>\n" );
document.write( "z ≈ -26.1 / 6.58<BR>\n" );
document.write( "z ≈ -3.97\r<BR>\n" );
document.write( "\n" );
document.write( "**4. P-value:**\r<BR>\n" );
document.write( "\n" );
document.write( "Because this is a left-tailed test, the p-value is the probability of getting a z-score as extreme as -3.97 or *lower*.  Using a z-table or calculator:\r<BR>\n" );
document.write( "\n" );
document.write( "P(z < -3.97) ≈ 0.0000 (very close to zero)\r<BR>\n" );
document.write( "\n" );
document.write( "**5. Decision:**\r<BR>\n" );
document.write( "\n" );
document.write( "Since the p-value (≈ 0.0000) is *less than* the significance level (α = 0.05), we *reject* the null hypothesis.\r<BR>\n" );
document.write( "\n" );
document.write( "**6. Conclusion:**\r<BR>\n" );
document.write( "\n" );
document.write( "There is sufficient evidence at the α = 0.05 level of significance to conclude that Internet users spend less time watching television than the average American.<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );